# Wikijunior:Kings and Queens of England/The House of Windsor

## George V (1910--1936)

The first monarch of the House of Windsor was George V. However, he
started out as being of the House of Saxe-Coburg-Gotha, and so he can be
read about in the previous chapter.

## Edward VIII (1936)

Edward VIII was born at White Lodge, Richmond upon Thames on 23 June
1894. Edward was the eldest son of George V and Queen Mary. He was King
of the United Kingdom of Great Britain and Northern Ireland and the
British Dominions beyond the Seas, and Emperor of India from 20 January
1936 until his abdication on 11 December 1936, after which he was Prince
Edward and then Duke of Windsor until his death on 28 May 1972. During
World War II he was the Governor and Commander-in-Chief of Bahamas after
spending a great deal of time in Bermuda. He was known to his family and
close friends, by his last Christian name, David.

### Prince of Wales

In 1910, George V made Edward Prince of Wales. When the First World War
broke out in 1914 Edward joined the army, serving with the Grenadier
Guards. Although Edward was willing to serve on the front lines, the
British government refused to allow it, because of the harm that the
capture of the heir to the throne would cause. However, Edward did
witness at first hand the horror of trench warfare, and visited the
front line as often as he could. His role in the war, although limited,
led to his great popularity among veterans of the conflict.

Throughout the 1920s the Prince of Wales represented his father, King
George V, at home and abroad on many occasions. He took a special
interest in visiting the poorest areas of the country. Abroad the Prince
of Wales toured the British Empire, undertaking 13 tours between 1919
and 1935.

In 1928, King George V gave Edward a home, Fort Belvedere, near
Sunningdale in Berkshire. There Edward had relationships with a series
of married women, including the American Wallis Simpson. Simpson had
divorced her first husband in 1927 and later married Ernest Simpson.
Edward\'s relationship with Wallis Simpson further weakened his poor
relationship with his father, King George V. The King and Queen refused
to receive Mrs Simpson at court, and his brother, Prince Albert, urged
Edward to seek a more suitable wife. Edward, however, had now fallen in
love with Wallis and the couple grew ever closer.

### Reign

King George V died on 20 January 1936, and Edward became King Edward
VIII. The next day he broke royal protocol by watching the proclamation
of his own accession to the throne from a window of St James\'s Palace
in the company of the still-married Mrs Simpson. It was also at this
time that Edward VIII became the first British monarch to fly in an
aeroplane, when he flew from Sandringham to London for his Accession
Council.

It was now becoming clear that the new King wished to marry Mrs Simpson,
especially when divorce proceedings between Mr and Mrs Simpson were
brought at Ipswich Crown Court. Powerful figures in the British
government deemed marriage to Mrs Simpson impossible for Edward, even if
Wallis obtained her second divorce, because he had become the Supreme
Governor of the Church of England which prohibited remarriage after
divorce.

On 16 July 1936, an attempt was made on the King\'s life. Jerome
Bannigan produced a loaded revolver as the King rode on horseback at
Constitution Hill, near Buckingham Palace. Police spotted the gun and
pounced on him, and he was quickly arrested. At Bannigan\'s trial, he
alleged that \"a foreign power\" had paid him £150 to kill Edward, a
claim the court rejected.

On 16 November 1936, Edward met with Prime Minister Stanley Baldwin at
Fort Belvedere and expressed his desire to marry Wallis Simpson when she
became free to do so. The Prime Minister responded by presenting the
King with three choices: he could give up the idea of marriage; marry
Wallis against his ministers\' wishes; or abdicate. It was clear that
Edward was not prepared to give up Wallis. By marrying against the
advice of his ministers, it was likely that he would cause the
government to resign, prompting a constitutional crisis. The Prime
Ministers of the British dominions had also made clear their opposition
to the King marrying a divorcée; only the Irish Free State was not
opposed to the idea of marriage. Faced with this opposition, Edward
chose to abdicate. Edward duly signed an instrument of abdication at
Fort Belvedere on 10 December 1936. The next day, he performed his last
act as King when he gave Royal Assent to His Majesty\'s Declaration of
Abdication Act 1936 which applied to the United Kingdom and all the
dominions except the Irish Free State. The Free State passed the
equivalent External Relations Act, which included the abdication in its
schedule, the next day.

On the night of 11 December 1936, Edward, now reverted to the title of
The Prince Edward, made a broadcast to the nation and the Empire,
explaining his decision to abdicate. He famously said, \"I have found it
impossible to carry the heavy burden of responsibility and to discharge
my duties as king as I would wish to do without the help and support of
the woman I love.\"

After the broadcast, Edward left the United Kingdom for France, though
he was unable to join Wallis until her divorce became absolute, several
months later. His brother, Prince Albert, Duke of York succeeded to the
throne as King George VI. George VI made Edward the Duke of Windsor. The
Duke of Windsor married Mrs Simpson, who had changed her name by deed
poll to Wallis Warfield, in a private ceremony on 3 June 1937 at Chateau
de Candé, Monts, France. None of the British royal family attended.

### World War II

In 1937, the Duke and Duchess visited Germany as personal guests of the
Nazi leader Adolf Hitler, a visit much publicised by the German media.
The couple then settled in France. When the Germans invaded the north of
France in May 1940, the Windsors fled south, first to Biarritz, then in
June to Spain. In July the pair moved to Lisbon. In August a British
warship took them to the Bahamas, where the Duke of Windsor was made
Governor. He held the post until the end of World War II in 1945.

### Death

After the war, the couple returned once again to France in Neuilly near
Paris, where they spent much of the remainder of their lives in
retirement. The Royal Family never accepted the Duchess and would not
receive her formally, although the former King sometimes met his mother
and a brother after his abdication. The Duke died of throat cancer in
1972 in Paris, and his body was returned to Britain for burial at
Frogmore near Windsor Castle. The Duchess travelled to England to attend
his funeral, staying at Buckingham Palace during her visit. The Duchess,
on her death a decade and a half later, was buried alongside her husband
in Frogmore simply as \"Wallis, his wife\". The Duke and Duchess had no
children.

## George VI (1936--1952)

!George VI{width="150"} George VI
was born at York Cottage on the Sandringham Estate in Norfolk on 14
December 1895. He was the second son of George V and Queen Mary. He was
King of the United Kingdom of Great Britain and Northern Ireland and the
British dominions beyond the seas from 11 December 1936 until his death
on 6 February 1952. George VI was also the Emperor of India (until 1947)
and King of Ireland (until 1949). George VI came the throne after the
abdication of his brother, Edward VIII. He was king during the Second
World War. He was known as Bertie, after his first name, Albert, to his
family.

### Early life

As a child, George often suffered from ill health and he was described
as \"easily frightened and somewhat prone to tears\". Albert developed a
severe stammer that lasted for many years as well as chronic stomach
problems. He also suffered from knock knees, and to correct this he had
to wear splints, which were extremely painful. He was also forced to
write with his right hand although he was a natural left-hander.

In 1909, Bertie joined the Royal Navy and served as a naval cadet.
Despite coming in at the bottom of the class, Albert moved to Dartmouth
and served as a midshipman. He was still in the Navy when Edward VII
died on 6 May 1910. Albert was now second in line for the throne. Albert
served during World War I. He saw action aboard HMS *Collingwood* in the
Battle of Jutland in 1916. In 1917 Albert joined the Royal Air Force but
did not see any further action in the war. After the war, Albert studied
history, economics and civics for a year at Trinity College, Cambridge,
from October 1919. In 1920, he began to take on royal duties,
representing his father, King George V.

### Marriage and children

Albert had a great deal of freedom in choosing his wife. In 1920 he met
Lady Elizabeth Bowes-Lyon, the youngest daughter of Claude Bowes-Lyon,
14th Earl of Strathmore and Kinghorne and set his sights on marrying
her. She rejected his proposal twice before she said yes. They were
married on 26 April 1923. They had two children: Elizabeth, born in
1926, and Margaret, born in 1930.

### Reign

The Duke and Duchess lived a relatively sheltered life at their London
residence, 145 Piccadilly. When Edward VIII he abdicated on 11 December
1936, however, Albert became King George VI.

The growing likelihood of war in Europe would dominate the reign of King
George VI. Initially the King and Queen took an appeasement stance
against the German dictator Adolf Hitler, and supported the policy of
Neville Chamberlain. In 1939, the King and Queen visited Canada from
where they made a shorter visit to the United States of America. George
was the first reigning British monarch to visit either of these
countries. The aim of the tour was mainly political, to shore up
Atlantic support for Britain in any upcoming war. The King and Queen
were extremely enthusiastically received by the Canadian public. They
were also warmly received by the American people, visiting the 1939 New
York World\'s Fair and staying at the White House with President
Roosevelt. When war broke out in 1939, George VI with his wife chose to
stay in London and not flee to Canada, as had been suggested. The King
and Queen officially stayed in Buckingham Palace throughout the war,
although they often escaped to Windsor Castle to avoid bombing raids.
George VI and Queen Elizabeth narrowly avoided death when a lone German
bomber despatched to bomb Buckingham Palace attacked. The bomb exploded
in the courtyard, shattering windows in the palace.

Throughout the war, the King and Queen provided morale-boosting visits
throughout the UK, visiting bomb sites and munition factories. On
Victory in Europe (VE) Day, the Royal Family appeared on the balcony of
Buckingham Palace to celebrate the end of the war in Europe. After the
war, George VI\'s reign saw the start of the break up of the British
Empire, in particular when India became an independent dominion, with
George VI giving up the title of Emperor of India.

### Death

The war had taken its toll on the King\'s health. This was made worse by
his heavy smoking and he got lung cancer. Increasingly his daughter
Princess Elizabeth, the heir to the throne, would take on more of the
royal duties as her father\'s health got worse. On 6 February 1952,
George VI died aged 56 in his sleep at Sandringham House in Norfolk. He
was buried in St George\'s Chapel in Windsor Castle. In 2002, the body
of his wife Elizabeth and the ashes of his daughter Princess Margaret
were interred in a tomb alongside him.

## Elizabeth II (1952--2022)

![](Elizabeth_II_greets_NASA_GSFC_employees,_May_8,_2007.jpg "Elizabeth_II_greets_NASA_GSFC_employees,_May_8,_2007.jpg")

Elizabeth II was born at 17 Bruton Street in Mayfair, London on 21 April
1926. She was the eldest daughter of George VI and Queen Elizabeth.
Elizabeth II was Queen of the United Kingdom of Great Britain and
Northern Ireland and of many other independent nations. She became Queen
on 6 February 1952 at age 25, and ruled until her death on 8 September
2022. Her 70-year reign was the second-longest after King Louis XIV of
France (72 years), and the longest of any British monarch ever.

### Early life

Elizabeth was thirteen years old when World War II broke out. She and
her younger sister, Princess Margaret, were evacuated to Windsor Castle,
Berkshire. There was some suggestion that the princesses be sent to
Canada, but their mother refused to consider this, famously saying,
\"The children could not possibly go without me, I wouldn\'t leave
without the King, and the King will never leave\". In 1940 Princess
Elizabeth made her first radio broadcast, addressing other children who
had been evacuated.

In 1945 Princess Elizabeth joined the Women\'s Auxiliary Territorial
Service. She was the first and so far only female member of the royal
family actually to serve in the military. Elizabeth made her first
official visit overseas in 1947, when she went with her parents to South
Africa. On her 21st birthday she made a broadcast to the British
Commonwealth and Empire pledging to devote her life to the service of
the people of the Commonwealth and Empire.

### Marriage and children

Elizabeth married Prince Philip of Greece and Denmark on 20 November
1947, who was created the Duke of Edinburgh on the same date. The Duke
was Queen Elizabeth\'s second cousin once removed, as they both
descended from Christian IX of Denmark. The couple were also third
cousins. After their wedding Philip and Elizabeth took up residence at
Clarence House, London. But at various times between 1946 and 1953, the
Duke of Edinburgh was stationed in Malta as a serving Royal Navy
officer. Philip and Elizabeth lived in Malta for a period between 1949
and 1951. On 14 November 1948 Elizabeth gave birth to her first child,
Charles. They later had three other children: Anne, Andrew and Edward.

### Reign

King George\'s health declined during 1951 and Elizabeth frequently
stood in for him at public events. She visited Greece, Italy and Malta
(where Philip was then stationed) during the year. In October she toured
Canada and visited President Truman in Washington, DC. In January 1952
Elizabeth and Philip set out for a tour of Australia and New Zealand.
They had reached Kenya when word arrived of the death of her father on 6
February 1952. They were staying at the Treetops Hotel, where Prince
Philip was the one who informed Elizabeth that she was now queen.

After the coronation, Elizabeth and Philip moved to Buckingham Palace in
central London. It is believed, however, that like many of her
predecessors she dislikes the Palace as a residence and considers
Windsor Castle, west of London, to be her home. She also spends time at
Balmoral Castle in Scotland and at Sandringham House in Norfolk.

At the time of Elizabeth\'s accession there was much talk of a \"new
Elizabethan age\". As nations have developed economically and in
literacy, Queen Elizabeth has witnessed over the past 50 years a gradual
transformation of the British Empire into its modern successor, the
Commonwealth. She has worked hard to maintain links with former British
possessions, and in some cases, such as South Africa after the end of
Apartheid, she has played an important role in retaining or restoring
good relations. Queen Elizabeth has become the most widely travelled
head of state in history and has visited many Commonwealth and
non-Commonwealth countries. Her son Charles, the Prince of Wales,
welcomed his first child Prince William in 1982.

In 2002, Queen Elizabeth celebrated her Golden Jubilee, marking the 50th
anniversary of her accession to the Throne. The year saw a large tour of
the Commonwealth Realms, including numerous parades and official
concerts. Then in 2012, the Queen celebrated her Diamond Jubilee,
marking the 60th anniversary of her accession to the Throne. A year
later, her grandson William had his first son Prince George. Queen
Elizabeth surpassed her great-great-grandmother Queen Victoria to become
the longest-reigning British monarch in 2015, and she later became the
first British monarch to celebrate her Sapphire Jubilee (65 years) and
Platinum Jubilee (70 years) on the throne.

In 2021, the Queen\'s husband Prince Philip, the Duke of Edinburgh, died
age 99, after 73 years of marriage. Her health started to decline around
the same time, and it became more common for the Queen to cancel royal
engagements. At the State Opening of Parliament in 2021, she could not
attend, and Prince Charles had to give the speech from the throne in her
place. Queen Elizabeth was also absent from the celebrations of her
Platinum Jubilee during the summer of 2022. Seven months later, on 8
September 2022, she died at Balmoral age 96, bringing the second
Elizabethan era to an end.

## Charles III (2022 onwards)

!Charles III, seen here in 2005 when he was still the Prince of
Wales.{width="160"}

Charles III was born at Buckingham Palace in London on 14 November 1948.
He is the eldest son of Elizabeth II and Prince Philip, the Duke of
Edinburgh. Charles III is King of the United Kingdom of Great Britain
and Northern Ireland and 14 other independent nations, as well as Head
of the Commonwealth. He became King on 8 September 2022 after the death
of his mother. He was only 3 years old when his mother became Queen, but
73 when she died. This makes Charles both the longest-serving heir to
the British throne, and the oldest new king in British history.

Before his accession, Charles (at the time known as the Prince of Wales)
worked extensively on charity, particularly for the Prince\'s Trust. His
personal life drew a large amount of public attention, particularly his
marriages to the late Lady Diana Spencer and later to Camilla, the
Duchess of Cornwall. In the final years of his mother\'s life, Charles
began taking over some of her royal duties.

### Early life

Charles was born when his grandfather George VI was on the throne, and
his mother Elizabeth was first in line. When George VI died in 1952, his
mother became Queen as Elizabeth II. Charles became first in line, and
also gained the titles Duke of Cornwall and Duke of Rothesay (in
Scotland only), which are titles that both automatically go to the heir
to the throne.

He went to Hill House School in West London, and later the Cheam
preparatory school in Berkshire, which the Duke of Edinburgh had also
gone to. The Prince finished his education at Gordonstoun, a private
boarding school in the north east of Scotland. His father, the Duke of
Edinburgh, had previously attended Gordonstoun, becoming head boy. It is
often reported that the Prince hated his time at the school, where he
was a frequent target for bullies. In 1966 Charles spent two terms at
Geelong Grammar School in Victoria, Australia during which time he also
visited Papua New Guinea. On his return to Gordonstoun he followed in
his father\'s footsteps by becoming head boy. In 1967 he left
Gordonstoun with two A levels, in history and French.

Traditionally, the heir to the throne would go straight into the
military after finishing school. However, Charles went to university at
Trinity College, Cambridge where he studied Anthropology and
Archaeology, and later History, earning a 2:2 (a lower second class
degree). Charles was the first member of the British royal family to be
awarded a degree. He also went to the University College of Wales,
Aberystwyth to learn the Welsh language, making him the first
English-born Prince of Wales ever to make a serious attempt to do so.

He was created the Prince of Wales and Earl of Chester in 1958, though
his actual investiture did not take place until 1 July 1969. This was a
major ceremony, held at Caernarfon Castle in north Wales, a place
traditionally associated with the creation of the title in the 13th
century. The Welsh borough of Swansea was granted city status to mark
the occasion. He served on HMS Norfolk as Sub Luietenant Prince of Wales
in the early seventies.

### Marriages and children

In 1980, the Prince of Wales began dating Lady Diana Spencer (born
1961), a woman from the Spencer family of nobility. They announced their
engagement on 24 February 1981, and their wedding took place at St
Paul\'s Cathedral on 29 July 1981, before 3,500 invited guests. An
estimated 750 million people watched it on television around the world,
making it one of the most viewed events in television history. Almost
immediately, Princess Diana became a star attraction, chased by the
press, her every move (including every change in hairstyle) closely
followed by millions. However, the marriage soon became troubled. Within
five years of the wedding the marriage was already on the brink of
collapse, and though they remained publicly a couple, they had
effectively separated by the mid-1990s, with him living in Highgrove and
her in Kensington Palace. They divorced on 28 August 1996. They had had
two sons, Prince William born in 1982 and Prince Henry (known by the
name \"Harry\") born two years later. Princess Diana died in a car
accident in Paris, France in 1997.

During Charles and Diana\'s marriage, and even before it, rumors were
circulating that Charles was having an affair with Camilla Parker Bowles
(herself married to British Army officer Andrew Parker Bowles). Prince
Charles confirmed this after his and Diana\'s divorce, explaining that
this was one reason for why their marriage fell apart. As Camilla and
her husband had also divorced, Charles and Camilla\'s relationship
became publicly known. Over time, opinion shifted to a point where a
civil marriage would be acceptable. They married on 9 April 2005 at
Windsor Guildhall in Berkshire.

Out of respect for Diana, who in the eyes of the public had become
closely associated with the title Princess of Wales, it was announced
that Camilla would not use this title (even though she as the wife of
the Prince of Wales was formally entitled to do so). Camilla was instead
Her Royal Highness the Duchess of Cornwall. It was also decided at the
time that when Prince Charles one day became king, Camilla would be
known as princess consort rather than queen. At the Platinum Jubilee of
Elizabeth II in 2022, the Queen announced that she wanted Camilla to be
known as queen when Charles became king, which happened when Elizabeth
II died later that year.

### Personal interests

Charles is a keen horseman and huntsman. He served in the Royal Navy
from 1971 to 1976, commanding HMS *Bronington*, a minehunter, from
February 1976 until December 1976. He is also a watercolour artist and a
published writer. He has exhibited and sold a number of paintings. The
Prince\'s Trust, which he founded, is a charity that works mainly with
young people, offering loans to groups, businesses and people (often in
deprived areas) who had difficulty receiving outside support.
Fundraising concerts are regularly held for the Prince\'s Trust, with
leading pop, rock and classical musicians taking part

To put his ideas on architecture and town planning into practice,
Charles is developing the village of Poundbury in Dorset. Prior to
commencing work on Poundbury he had published a book and produced a
documentary called *A Vision for Britain*. In 1992 he also established
The Prince of Wales Institute of Architecture and began the publication
of a magazine dealing with architecture, but the latter has since ceased
independent operation after being merged with another charity to create
The Prince\'s Foundation for the Built Environment in 2001. He is also
keen on growing and promoting organic food, although he drew some
ridicule when he joked about sometimes talking to his houseplants.
Before becoming king, Charles was president of 16 charities, and raised
over £100 million for charity in 2004.

### Reign

Charles became king immediately when his mother died on 8 September
2022. The following day, he named his eldest son William as the new
Prince of Wales. Charles III and his wife Camilla were coronated on 6
May 2023 in a ceremony at Westminster Abbey.


# Wikijunior:Kings and Queens of England/Future monarchs

## Prince William of Wales

!Prince
William{width="150"}
Prince William of Wales was born at St. Mary\'s Hospital, Paddington on
21 June 1982. His parents are Charles, Prince of Wales and Lady Diana
Spencer. He is next in the line of succession to the British throne.

On 1 March 1991 (St. David\'s Day), Prince William made his first
official public appearance during a visit to Cardiff, the capital city
of Wales. After arriving by plane, the eight-year-old prince was taken
by his parents to Llandaff Cathedral. After a tour of the cathedral, he
signed its visitors book, which showed that he was left-handed. Prince
William went to the Mrs Jane Mynors\' nursery school and the pre-prep
Wetherby School both in West London, and later attended Ludgrove School
in Berkshire. After passing an entrance exam he went on to attend Eton
College in Berkshire. Whilst there he studied geography, biology and
history of art at A-level.

Like a growing number of British teenagers, Prince William chose to take
a gap year after finishing Eton College. He took part in British Army
training in Belize. He spent the final stage of his gap year in southern
Chile as a volunteer with Raleigh International. Pictures of the prince
cleaning a toilet were broadcast around the world. After his gap year,
Prince William attended the University of St Andrews in Fife, Scotland
from 2001 until he was graduated in 2005. He started doing a degree
course in Arts History, although he later changed his main subject to
Geography and gained a Scottish Master of Arts degree with upper-second
class honours. In January 2006, the prince began his cadet course at the
Royal Military Academy Sandhurst to train as an Army Officer. William
joined his brother who had been there since May 2005.

In July 2005, William carried out his first official engagements
representing Elizabeth II, as Queen of New Zealand, at World War II
commemorations in New Zealand. In September of the same year, it was
announced that William will become president of The Football Association
in May 2006 and patron to the UK charity Centrepoint, which works with
homeless young people. In April 2011, he married Kate Middleton. In July
2013, the Duchess of Cambridge gave birth to their son, Prince George of
Cambridge, third in line to the throne. The couple\'s second child,
their daughter Princess Charlotte of Cambridge who is fourth in line to
the throne, was born in May 2015. Prince Louis of Cambridge, the fifth
in line to the throne, was born in April 2018.

## Prince George

!Prince George (2013).jpg "Prince George (2013)")
After William, Prince George of Cambridge will be the king of England.
He was born on the 22nd of July, 2013. His full name is George Alexander
Louis.


# Wikijunior:Languages/Playing with language

We\'ve looked at lots of existing languages from all around the world,
but if we really want to have fun with languages then we have to do more
than just look; we have to play as well. Imagine if you could talk to
your friends in your own private language that no one else knew. Or
imagine speaking a language where everything you said sounded silly, or
grumpy, or smart. Or imagine making up your own special words for all
the things you see around you. In this chapter we\'ll find out how to do
all these things and more.

Languages are like the recipes for cakes. You start off with a whole
bunch of different ingredients and then the recipe tells you how to mix
them all together. In a cake you might use some flour and some eggs, a
little bit of milk and butter and then maybe some chocolate. In a
language you use sounds and words and when you mix them all together you
end up with a sentence. And just like in a cake it is important how you
combine the ingredients. If you put too much water in your cake you will
end up with a slimy sludge instead of a cake. In a language if you
don\'t put your words in the right order, if you don\'t follow the
language\'s recipe, then you end up with a slimy sludge too. Let\'s try
it with a simple sentence:

If we didn\'t follow the recipe we might get:

```{=html}
<div class="center">
```
*The moon is made of bright green cheese.*

*Cheese is made of the bright green moon.*

*Green made of the moon is cheese bright.*

```{=html}
</div>
```
And these are just two of the silly sentences you can make by not
following the recipe for the language correctly. Try some more by
yourself, for example you could try saying it in reverse
order---starting with the last word and going to the first. No matter
how you rearrange the words, there is only one correct recipe that says
that the moon is made of bright green cheese. Some of the other recipes
make sentences that make sense but don\'t say what we want them to say,
but most recipes just end up being silly. The \"recipe\" of a language
that determine which words go where, and what they mean is the *syntax*.

Without even realizing it, you already know the recipe to one
language---English. You know that putting words in some orders makes
sense but in others it doesn\'t. But imagine if you only ever had one
kind of cake. It would get boring. It\'s the same in languages: the more
different language recipes you know, the more fun it is.

If you wanted to make a language recipe there are two different ways you
could do it. The first is to get an existing language recipe (for
example, English) and use it to make a language cake, putting lots of
icing on it so that nobody other than you and your friends know that
underneath the icing it is really the same old cake. The other is to
make a whole new recipe from scratch. Because it is easier to add icing
to a cake that already exists rather than bake a whole new one, we\'ll
start with that.

## Putting the icing on our language cake

*Nac uoy daer siht ecnetnes? eno siht tuoba woH?*

If you can read English then you can read these sentences -- but you
might need a hint. When we write a normal word, we start at the left
side of a word and keep writing until we reach the right side. But what
if we started on the right side and wrote to the left---what if we wrote
words backwards? Try reading the sentences above again, but this time
read all the words backwards. You might find it easier if you try
writing them down.

So what did that say? Did it make sense? Try reading the second sentence
again but this time instead of reading each word backwards read the
whole sentence backwards. Now try saying it out loud. What does it sound
like? Is it difficult to say? Why are some of the words hard to say but
some of them are easy? We\'ll find out in a later section .

You can use talking or writing backwards as a secret language for you
and your friends.

## Making a language recipe from scratch

Remember the chapter on
Quenya? Well, it was a
language recipe that J.R.R. Tolkien made up. He spent a large part of
his life making Quenya and never really finished it, but don\'t worry:
while the language recipes we make here will not be complete like real
languages, they will still be usable, so you will be able to say simple
things to your friends with them, without anybody else knowing what you
are saying.

### How to make a language sound the way you want it to

### All the amazing things words can do

### Mixing all your words together to make sentences


# Linear Algebra/Introduction

|current=Introduction
|previous=Notation
book helps students to master the material of a standard undergraduate
linear algebra course.

The material is standard in that the topics covered are Gaussian
reduction, vector spaces, linear maps, determinants, and eigenvalues and
eigenvectors. The audience is also standard: sophomores or juniors,
usually with a background of at least one semester of calculus and
perhaps with as many as three semesters.

The help that it gives to students comes from taking a developmental
approach---this book\'s presentation emphasizes motivation and
naturalness, driven home by a wide variety of examples and extensive,
careful, exercises. The developmental approach is what sets this book
apart, so some expansion of the term is appropriate here.

Courses in the beginning of most mathematics programs reward students
less for understanding the theory and more for correctly applying
formulas and algorithms. Later courses ask for mathematical maturity:
the ability to follow different types of arguments, a familiarity with
the themes that underlie many mathematical investigations like
elementary set and function facts, and a capacity for some independent
reading and thinking. Linear algebra is an ideal spot to work on the
transition between the two kinds of courses. It comes early in a program
so that progress made here pays off later, but also comes late enough
that students are often majors and minors. The material is coherent,
accessible, and elegant. There are a variety of argument styles---proofs
by contradiction, if and only if statements, and proofs by induction,
for instance---and examples are plentiful.

So, the aim of this book\'s exposition is to help students develop from
being successful at their present level, in classes where a majority of
the members are interested mainly in applications in science or
engineering, to being successful at the next level, that of serious
students of the subject of mathematics itself.

Helping students make this transition means taking the mathematics
seriously, so all of the results in this book are proved. On the other
hand, we cannot assume that students have already arrived, and so in
contrast with more abstract texts, we give many examples and they are
often quite detailed.

In the past, linear algebra texts commonly made this transition
abruptly. They began with extensive computations of linear systems,
matrix multiplications, and determinants. When the concepts---vector
spaces and linear maps---finally appeared, and definitions and proofs
started, often the change brought students to a stop. In this book,
while we start with a computational topic, linear reduction, from the
first we do more than compute. We do linear systems quickly but
completely, including the proofs needed to justify what we are
computing. Then, with the linear systems work as motivation and at a
point where the study of linear combinations seems natural, the second
chapter starts with the definition of a real vector space. This occurs
by the end of the third week.

Another example of our emphasis on motivation and naturalness is that
the third chapter on linear maps does not begin with the definition of
homomorphism, but with that of isomorphism. That\'s because this
definition is easily motivated by the observation that some spaces are
\"just like\" others. After that, the next section takes the reasonable
step of defining homomorphism by isolating the operation-preservation
idea. This approach loses mathematical slickness, but it is a good trade
because it comes in return for a large gain in sensibility to students.

One aim of a developmental approach is that students should feel
throughout the presentation that they can see how the ideas arise, and
perhaps picture themselves doing the same type of work.

The clearest example of the developmental approach taken here---and the
feature that most recommends this book---is the exercises. A student
progresses most while doing the exercises, so they have been selected
with great care. Each problem set ranges from simple checks to
reasonably involved proofs. Since an instructor usually assigns about a
dozen exercises after each lecture, each section ends with about twice
that many, thereby providing a selection. There are even a few problems
that are challenging puzzles taken from various journals, competitions,
or problems collections. (These are marked with a \"**?**\" and as part
of the fun, the original wording has been retained as much as possible.)
In total, the exercises are aimed to both build an ability at, and help
students experience the pleasure of, *doing* mathematics.

## Applications and Computers.

The point of view taken here, that linear algebra is about vector spaces
and linear maps, is not taken to the complete exclusion of others.
Applications and the role of the computer are important and vital
aspects of the subject. Consequently, each of this book\'s chapters
closes with a few application or computer-related topics. Some are:
network flows, the speed and accuracy of computer linear reductions,
Leontief Input/Output analysis, dimensional analysis, Markov chains,
voting paradoxes, analytic projective geometry, and difference
equations.

These topics are brief enough to be done in a day\'s class or to be
given as independent projects for individuals or small groups. Most
simply give the reader a taste of the subject, discuss how linear
algebra comes in, point to some further reading, and give a few
exercises. In short, these topics invite readers to see for themselves
that linear algebra is a tool that a professional must have.

## For people reading this book on their own.

This book\'s emphasis on motivation and development make it a good
choice for self-study. But, while a professional instructor can judge
what pace and topics suit a class, if you are an independent student
then perhaps you would find some advice helpful.

Here are two timetables for a semester. The first focuses on core
material.

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calculus).

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See the table of contents for the titles of these subsections.

To help you make time trade-offs, in the table of contents I have marked
subsections as optional if some instructors will pass over them in favor
of spending more time elsewhere. You might also try picking one or two
topics that appeal to you from the end of each chapter. You\'ll get more
from these if you have access to computer software that can do the big
calculations.

The most important advice is: do many exercises. The recommended
exercises are labeled throughout. (The answers are available.) You
should be aware, however, that few inexperienced people can write
correct proofs. Try to find a knowledgeable person to work with you on
this.

Finally, if I may, a caution for all students, independent or not: I
cannot overemphasize how much the statement that I sometimes hear, \"I
understand the material, but it\'s only that I have trouble with the
problems\" reveals a lack of understanding of what we are up to. Being
able to do things with the ideas is their point. The quotes below
express this sentiment admirably. They state what I believe is the key
to both the beauty and the power of mathematics and the sciences in
general, and of linear algebra in particular (I took the liberty of
formatting them as poems).

*I know of no better tactic*\
* than the illustration of exciting principles*\
*by well-chosen particulars.*\
        *\--Stephen Jay Gould*

*If you really wish to learn*\
* then you must mount the machine*\
* and become acquainted with its tricks*\
*by actual trial.*\
        *\--Wilbur Wright*

Jim Hefferon\
Mathematics, Saint Michael\'s College\
Colchester, Vermont USA 05439\
`http://joshua.smcvt.edu`\
2006-May-20

*Author\'s Note.* Inventing a good exercise, one that enlightens as well
as tests, is a creative act, and hard work.

The inventor deserves recognition. But for some reason texts have
traditionally not given attributions for questions. I have changed that
here where I was sure of the source. I would greatly appreciate hearing
from anyone who can help me to correctly attribute others of the
questions.

ja:線型代数学/序論


# Linear Algebra/Gauss' Method

|current=Gauss' Method
|previous=Solving Linear Systems

Finding the set of all solutions is
**solving**
the system. No guesswork or good fortune is needed to solve a linear
system. There is an algorithm that always works. The next example
introduces that algorithm, called **Gauss\'
method**. It transforms the
system, step by step, into one with a form that is easily solved.

Most of this subsection and the next one consists of examples of solving
linear systems by Gauss\' method. We will use it throughout this book.
It is fast and easy. But, before we get to those examples, we will first
show that this method is also safe in that it never loses solutions or
picks up extraneous solutions.

Each of those three operations has a restriction. Multiplying a row by
$0$ is not allowed because obviously that can change the solution set of
the system. Similarly, adding a multiple of a row to itself is not
allowed because adding $-1$ times the row to itself has the effect of
multiplying the row by $0$. Finally, swapping a row with itself is
disallowed to make some results in the fourth chapter easier to state
and remember (and besides, self-swapping doesn\'t accomplish anything).

When writing out the calculations, we will abbreviate \"row $i$\" by
\"$\rho_i$\". For instance, we will denote a pivot operation by
$k\rho_i+\rho_j$, with the row that is changed written second. We will
also, to save writing, often list pivot steps together when they use the
same $\rho_i$.

As these examples illustrate, Gauss\' method uses the elementary
reduction operations to set up back-substitution.

Strictly speaking, the operation of rescaling rows is not needed to
solve linear systems. We have included it because we will use it later
in this chapter as part of a variation on Gauss\' method, the
Gauss-Jordan method.

All of the systems seen so far have the same number of equations as
unknowns. All of them have a solution, and for all of them there is only
one solution. We finish this subsection by seeing for contrast some
other things that can happen.

That example\'s system has more equations than variables. Gauss\' method
is also useful on systems with more variables than equations. Many
examples are in the next subsection.

Another way that linear systems can differ from the examples shown
earlier is that some linear systems do not have a unique solution. This
can happen in two ways.

The first is that it can fail to have any solution at all.

The other way that a linear system can fail to have a unique solution is
to have many solutions.

Don\'t be fooled by the \"$0=0$\" equation in that example. It is not
the signal that a system has many solutions.

We will finish this subsection with a summary of what we\'ve seen so far
about Gauss\' method.

Gauss\' method uses the three row operations to set a system up for back
substitution. If any step shows a contradictory equation then we can
stop with the conclusion that the system has no solutions. If we reach
echelon form without a contradictory equation, and each variable is a
leading variable in its row, then the system has a unique solution and
we find it by back substitution. Finally, if we reach echelon form
without a contradictory equation, and there is not a unique solution (at
least one variable is not a leading variable) then the system has many
solutions.

The next subsection deals with the third case--- we will see how to
describe the solution set of a system with many solutions.

## Exercises

## References

-       .

-       .

-       .

-       .

-       .


# Linear Algebra/Describing the Solution Set

|current=Describing the Solution Set
|previous=Gauss' Method

A linear system with a unique solution has a solution set with one
element. A linear system with no solution has a solution set that is
empty. In these cases the solution set is easy to describe. Solution
sets are a challenge to describe only when they contain many elements.

In the echelon form system derived in the above example, $x$ and $y$ are
leading variables and $z$ is free.

We refer to a variable used to describe a family of solutions as a
**parameter** and we say that the set above is **parametrized** with $y$
and $w$ . (The terms \"parameter\" and \"free variable\" do not mean the
same thing. Above, $y$ and $w$ are free because in the echelon form
system they do not lead any row. They are parameters because they are
used in the solution set description. We could have instead parametrized
with $y$ and $z$ by rewriting the second equation as
$w=\frac23-\frac13z$ . In that case, the free variables are still $y$
and $w$ , but the parameters are $y$ and $z$ . Notice that we could not
have parametrized with $x$ and $y$ , so there is sometimes a restriction
on the choice of parameters. The terms \"parameter\" and \"free\" are
related because, as we shall show later in this chapter, the solution
set of a system can always be parametrized with the free variables.
Consequently, we shall parametrize all of our descriptions in this way.)

We finish this subsection by developing the notation for linear systems
and their solution sets that we shall use in the rest of this book.

Matrices are usually named by upper case roman letters, e.g. $A$ . Each
entry is denoted by the corresponding lower-case letter, e.g. $a_{i,j}$
is the number in row $i$ and column $j$ of the array. For instance,

$$A=\begin{pmatrix}1&2.2&5\\3&4&-7\end{pmatrix}$$

has two rows and three columns, and so is a $2\times3$ matrix. (Read
that \"two-by-three\"; the number of rows is always stated first.) The
entry in the second row and first column is $a_{2,1}=3$ . Note that the
order of the subscripts matters: $a_{1,2}\ne a_{2,1}$ since
$a_{1,2}=2.2$ . (The parentheses around the array are a typographic
device so that when two matrices are side by side we can tell where one
ends and the other starts.)

Matrices occur throughout this book. We shall use
$\mathcal{M}_{n\times m}$ to denote the collection of $n\times m$
matrices.

We will also use the array notation to clarify the descriptions of
solution sets. A description like
$\{(2-2z+2w,-1+z-w,z,w)\big|z,w\in\R\}$ from Example
2.3 is hard to read. We will rewrite it to
group all the constants together, all the coefficients of $z$ together,
and all the coefficients of $w$ together. We will write them vertically,
in one-column wide matrices.

$$\left\{\begin{pmatrix}2\\-1\\0\\0\end{pmatrix}+\begin{pmatrix}-2\\1\\1\\0\end{pmatrix}z+\begin{pmatrix}2\\-1\\0\\1\end{pmatrix}w\Bigg|z,w\in\R\right\}$$

For instance, the top line says that $x=2-2z+2w$ . The next section
gives a geometric interpretation that will help us picture the solution
sets when they are written in this way.

Vectors are an exception to the convention of representing matrices with
capital roman letters. We use lower-case roman or greek letters
overlined with an arrow: $\vec a,\vec b$ \... or
$\vec{\alpha},\vec{\beta}$ \... (boldface is also common: $\mathbf{a}$
or $\boldsymbol{\alpha}$). For instance, this is a column vector with a
third component of $7$ .

$$\vec v=\begin{pmatrix}1\\3\\7\end{pmatrix}$$

The style of description of solution sets that we use involves adding
the vectors, and also multiplying them by real numbers, such as the $z$
and $w$ . We need to define these operations.

Scalar multiplication can be written in either order: $r\cdot\vec v$ or
$\vec v\cdot r$ , or without the \"$\cdot$\" symbol: $r\vec v$ . (Do not
refer to scalar multiplication as \"scalar product\" because that name
is used for a different operation.)

Notice that the definitions of vector addition and scalar multiplication
agree where they overlap, for instance, $\vec v+\vec v=2\vec v$ .

With the notation defined, we can now solve systems in the way that we
will use throughout this book.

Before the exercises, we pause to point out some things that we have yet
to do.

The first two subsections have been on the mechanics of Gauss\' method.
Except for one result, Theorem
1.4 --- without
which developing the method doesn\'t make sense since it says that the
method gives the right answers--- we have not stopped to consider any of
the interesting questions that arise.

For example, can we always describe solution sets as above, with a
particular solution vector added to an unrestricted linear combination
of some other vectors? The solution sets we described with unrestricted
parameters were easily seen to have infinitely many solutions so an
answer to this question could tell us something about the size of
solution sets. An answer to that question could also help us picture the
solution sets, in $\mathbb{R}^2$, or in $\mathbb{R}^3$, etc.

Many questions arise from the observation that Gauss\' method can be
done in more than one way (for instance, when swapping rows, we may have
a choice of which row to swap with). Theorem
1.4 says that
we must get the same solution set no matter how we proceed, but if we do
Gauss\' method in two different ways must we get the same number of free
variables both times, so that any two solution set descriptions have the
same number of parameters? Must those be the same variables (e.g., is it
impossible to solve a problem one way and get $y$ and $w$ free or solve
it another way and get $y$ and $z$ free)?

In the rest of this chapter we answer these questions. The answer to
each is \"yes\". The first question is answered in the last subsection
of this section. In the second section we give a geometric description
of solution sets. In the final section of this chapter we tackle the
last set of questions. Consequently, by the end of the first chapter we
will not only have a solid grounding in the practice of Gauss\' method,
we will also have a solid grounding in the theory. We will be sure of
what can and cannot happen in a reduction.

## Exercises

## References

-       *The USSR Mathematics Olympiad*, number 174.

-


# Linear Algebra/General = Particular + Homogeneous

|current=General = Particular + Homogeneous
|previous=Describing the Solution Set

## Description of Solution Sets

The prior subsection has many descriptions of solution sets. They all
fit a pattern. They have a vector that is a particular solution of the
system added to an unrestricted combination of some other vectors. The
solution set from Example
2.13
illustrates.

$$\left\{
\underbrace{
\begin{pmatrix} 0 \\ 4 \\ 0 \\ 0 \\ 0 \end{pmatrix}}_{\begin{array}{c}\\[-19pt]\scriptstyle\text{particular} \\[-5pt]\scriptstyle\text{solution}\end{array}}+
\underbrace{w\begin{pmatrix} 1 \\ -1 \\ 3 \\ 1 \\ 0 \end{pmatrix}+
u\begin{pmatrix} 1/2 \\ -1 \\ 1/2 \\ 0 \\ 1 \end{pmatrix}}_{\begin{array}{c}\\[-19pt]\scriptstyle\text{unrestricted}\\[-5pt]\scriptstyle\text{combination}\end{array}}
\,\big|\, w,u\in\mathbb{R}\right\}$$

The combination is unrestricted in that $w$ and $u$ can be any real
numbers--- there is no condition like \"such that $2w-u=0$\" that would
restrict which pairs $w,u$ can be used to form combinations.

That example shows an infinite solution set conforming to the pattern.
We can think of the other two kinds of solution sets as also fitting the
same pattern. A one-element solution set fits in that it has a
particular solution, and the unrestricted combination part is a trivial
sum (that is, instead of being a combination of two vectors, as above,
or a combination of one vector, it is a combination of no vectors). A
zero-element solution set fits the pattern since there is no particular
solution, and so the set of sums of that form is empty.

We will show that the examples from the prior subsection are
representative, in that the description pattern discussed above holds
for every solution set.

This description has two parts, the particular solution $\vec{p}$ and
also the unrestricted linear combination of the $\vec{\beta}$\'s. We
shall prove the theorem in two corresponding parts, with two lemmas.

## Homogeneous Systems

We will focus first on the unrestricted combination part. To do that, we
consider systems that have the vector of zeroes as one of the particular
solutions, so that $\vec{p}+c_1\vec{\beta}_1+\dots+c_k\vec{\beta}_k$ can
be shortened to $c_1\vec{\beta}_1+\dots+c_k\vec{\beta}_k$.

(These are \"homogeneous\" because all of the terms involve the same
power of their variable--- the first power--- including a \"$0x_{0}$\"
that we can imagine is on the right side.)

Studying the associated homogeneous system has a great advantage over
studying the original system. Nonhomogeneous systems can be
inconsistent. But a homogeneous system must be consistent since there is
always at least one solution, the vector of zeros.

There are many different zero vectors, e.g., the one-tall zero vector,
the two-tall zero vector, etc. Nonetheless, people often refer to
\"the\" zero vector, expecting that the size of the one being discussed
will be clear from the context.

We now have the terminology to prove the two parts of Theorem
3.1 . The first lemma deals with
unrestricted combinations.

Before the proof, we will recall the back substitution calculations that
were done in the prior subsection.

Imagine that we have brought a system to this echelon form.

$$\begin{array}{*{4}{rc}r}
x  &+  &2y  &-  &z  &+  &2w &=  &0  \\
&   &-3y &+  &z  &   &   &=  &0  \\
&   &    &   &   &   &-w &=  &0
\end{array}$$

We next perform back-substitution to express each variable in terms of
the free variable $z$. Working from the bottom up, we get first that $w$
is $0\cdot z$, next that $y$ is $(1/3)\cdot z$, and then substituting
those two into the top equation $x+2((1/3)z)-z+2(0)=0$ gives
$x=(1/3)\cdot z$. So, back substitution gives a parametrization of the
solution set by starting at the bottom equation and using the free
variables as the parameters to work row-by-row to the top. The proof
below follows this pattern.

*Comment:* That is, this proof just does a verification of the
bookkeeping in back substitution to show that we haven\'t overlooked any
obscure cases where this procedure fails, say, by leading to a division
by zero. So this argument, while quite detailed, doesn\'t give us any
new insights. Nevertheless, we have written it out for two reasons. The
first reason is that we need the result--- the computational procedure
that we employ must be verified to work as promised.
The second reason is that the
row-by-row nature of back substitution leads to a proof that uses the
technique of mathematical induction.[^1] This is an important, and
non-obvious, proof technique that we shall use a number of times in this
book. Doing an induction argument here gives us a chance to see one in a
setting where the proof material is easy to follow, and so the technique
can be studied. Readers who are unfamiliar with induction arguments
should be sure to master this one and the ones later in this chapter
before going on to the second chapter.

`</math>`{=html} and substitute its expression in terms of free
variables. The result has the form

$$a_{m-(t+1),\ell_{m-(t+1)}}x_{\ell_{m-(t+1)}}+
\text{sums of multiples of free variables}=0$$

where $a_{m-(t+1),\ell_{m-(t+1)}}\neq 0$. We move the free variables to
the right-hand side and divide by $a_{m-(t+1),\ell_{m-(t+1)}}$, to end
with $x_{\ell_{m-(t+1)}}$ expressed in terms of free variables.

Because we have shown both the base step and the inductive step, by the
principle of mathematical induction the proposition is true. }}

We say that the set
$\{c_1\vec{\beta}_1+\cdots+c_k\vec{\beta}_k \,\big|\, c_1,\ldots,c_k\in\mathbb{R}\}$
is **generated by** or **spanned by** the set of vectors
$\{{\vec{\beta}_1},\ldots,{\vec{\beta}_k}\}$. There is a tricky point to
this definition. If a homogeneous system has a unique solution, the zero
vector, then we say the solution set is generated by the empty set of
vectors. This fits with the pattern of the other solution sets: in the
proof above the solution set is derived by taking the $c$\'s to be the
free variables and if there is a unique solution then there are no free
variables.

This proof incidentally shows, as discussed after Example
2.4
, that solution sets can always be parametrized using the free
variables.

## Nonhomogeneous Systems

The next lemma finishes the proof of Theorem
3.1 by considering the particular
solution part of the solution set\'s description.

The two lemmas above together establish Theorem
3.1 . We remember that theorem with
the slogan
\"$\text{General} = \text{Particular} + \text{Homogeneous}$\".

This table summarizes the factors affecting the size of a general
solution.

+---------------+-----------+---------------------------------+-------------------+
|               |           | *number of solutions of the*\   |                   |
|               |           | *associated homogeneous system* |                   |
+---------------+-----------+---------------------------------+-------------------+
|               |           | *one*                           | *infinitely many* |
+---------------+-----------+---------------------------------+-------------------+
| *particular*\ | *yes*     | unique\                         | infinitely many\  |
| *solution*\   |           | solution                        | solutions         |
| *exists?*     |           |                                 |                   |
+---------------+-----------+---------------------------------+-------------------+
| *no*          | no\       | no\                             |                   |
|               | solutions | solutions                       |                   |
+---------------+-----------+---------------------------------+-------------------+

The factor on the top of the table is the simpler one. When we perform
Gauss\' method on a linear system, ignoring the constants on the right
side and so paying attention only to the coefficients on the left-hand
side, we either end with every variable leading some row or else we find
that some variable does not lead a row, that is, that some variable is
free. (Of course, \"ignoring the constants on the right\" is formalized
by considering the associated homogeneous system. We are simply putting
aside for the moment the possibility of a contradictory equation.)

A nice insight into the factor on the top of this table at work comes
from considering the case of a system having the same number of
equations as variables. This system will have a solution, and the
solution will be unique, if and only if it reduces to an echelon form
system where every variable leads its row, which will happen if and only
if the associated homogeneous system has a unique solution. Thus, the
question of uniqueness of solution is especially interesting when the
system has the same number of equations as variables.

The above table has two factors. We have already considered the factor
along the top: we can tell which column a given linear system goes in
solely by considering the system\'s left-hand side--- the constants on
the right-hand side play no role in this factor. The table\'s other
factor, determining whether a particular solution exists, is tougher.
Consider these two

$$\begin{array}{*{2}{rc}r}
3x &+ &2y &= &5  \\
3x &+ &2y &= &5
\end{array}
\qquad
\begin{array}{*{2}{rc}r}
3x &+ &2y &= &5  \\
3x &+ &2y &= &4
\end{array}$$

with the same left sides but different right sides. Obviously, the first
has a solution while the second does not, so here the constants on the
right side decide if the system has a solution. We could conjecture that
the left side of a linear system determines the number of solutions
while the right side determines if solutions exist, but that guess is
not correct. Compare these two systems

$$\begin{array}{*{2}{rc}r}
3x &+ &2y &= &5  \\
4x &+ &2y &= &4
\end{array}
\qquad
\begin{array}{*{2}{rc}r}
3x &+ &2y &= &5  \\
3x &+ &2y &= &4
\end{array}$$

with the same right sides but different left sides. The first has a
solution but the second does not. Thus the constants on the right side
of a system don\'t decide alone whether a solution exists; rather, it
depends on some interaction between the left and right sides.

For some intuition about that interaction, consider this system with one
of the coefficients left as the parameter $c$.

$$\begin{array}{*{3}{rc}r}
x  &+  &2y  &+  &3z  &=  &1  \\
x  &+  &y   &+  &z   &=  &1  \\
cx  &+  &3y  &+  &4z  &=  &0
\end{array}$$

If $c=2$ this system has no solution because the left-hand side has the
third row as a sum of the first two, while the right-hand does not. If
$c\neq 2$ this system has a unique solution (try it with $c=1$). For a
system to have a solution, if one row of the matrix of coefficients on
the left is a linear combination of other rows, then on the right the
constant from that row must be the same combination of constants from
the same rows.

More intuition about the interaction comes from studying linear
combinations. That will be our focus in the second chapter, after we
finish the study of Gauss\' method itself in the rest of this chapter.

## Exercises

## Footnotes

```{=html}
<references/>
```
|current=General = Particular + Homogeneous
|previous=Describing the Solution Set
|next=Comparing Set Descriptions}}`{=mediawiki}

[^1]: More information on mathematical induction is in the appendix.


# Linear Algebra/Comparing Set Descriptions

|current=Comparing Set Descriptions
|previous=General = Particular + Homogeneous

*This subsection is optional. Later material will not require the work
here.*

## Comparing Set Descriptions

A set can be described in many different ways. Here are two different
descriptions of a single set:

$$\{\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}z\,\big|\, z\in\mathbb{R}\}
\quad\text{and}\quad
\{\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}w\,\big|\, w\in\mathbb{R}\}.$$

For instance, this set contains

$$\begin{pmatrix} 5 \\ 10 \\ 15 \end{pmatrix}$$

(take $z=5$ and $w=5/2$) but does not contain

$$\begin{pmatrix} 4 \\ 8 \\ 11 \end{pmatrix}$$

(the first component gives $z=4$ but that clashes with the third
component, similarly the first component gives $w=4/5$ but the third
component gives something different). Here is a third description of the
same set:

$$\{\begin{pmatrix} 3 \\ 6 \\ 9 \end{pmatrix}+\begin{pmatrix} -1 \\ -2 \\ -3 \end{pmatrix}y\,\big|\, y\in\mathbb{R}\}.$$

We need to decide when two descriptions are describing the same set.
More pragmatically stated, how can a person tell when an answer to a
homework question describes the same set as the one described in the
back of the book?

## Set Equality

Sets are equal if and only if they have the same members.
A common way to show that two
sets, $S_1$ and $S_2$, are equal is to show mutual inclusion: any member
of $S_1$ is also in $S_2$, and any member of $S_2$ is also in $S_1$.[^1]

## Exercises

## Footnotes

```{=html}
<references/>
```
|current=Comparing Set Descriptions
|previous=General = Particular + Homogeneous
|next=Automation}}`{=mediawiki}

[^1]: More information on set equality is in the appendix.


# Linear Algebra/Automation

|current=Automation
|previous=Comparing Set Descriptions

This is a PASCAL routine to do $k\rho_i+\rho_j$ to an augmented matrix.

``` pascal
PROCEDURE Pivot(VAR LinSys : AugMat;
k : REAL;
i, j : INTEGER)
VAR
Col : INTEGER;
BEGIN
FOR Col:=1 TO NumVars+1 DO
LinSys[j,Col]:=k*LinSys[i,Col]+LinSys[j,Col];
END;
```

Of course this is only one part of a whole program, but it makes the
point that Gaussian reduction is ideal for computer coding.

There are pitfalls, however. For example, some arise from the
computer\'s use of finite-precision approximations of real numbers.

These systems provide a simple example.
![](Linalg_singular_and_nonsingular_systems.png "Linalg_singular_and_nonsingular_systems.png")
(The second two lines are hard to tell apart.) Both have $(1,1)$ as
their unique solution.

In the first system, some small change in the numbers will produce only
a small change in the solution:

$$\begin{array}{*{2}{rc}r}
x  &+  &2y &= &3                  \\
3x &-  &2y &= &1.008
\end{array}$$

gives a solution of $(1.002,0.999)$. Geometrically, changing one of the
lines by a small amount does not change the intersection point by very
much.

That\'s not true for the second system. A small change in the
coefficients

$$\begin{array}{*{2}{rc}r}
x              &+ &2y  &=  &3                  \\
1.000\,000\,01x &+ &2y  &=  &3.000\,000\,03
\end{array}$$

leads to a completely different answer: $(3,0)$.

The solution of the second example varies wildly, depending on a $9$-th
digit. That\'s bad news for a machine using $8$ digits to represent
reals. In short, systems that are nearly singular may be hard to compute
with.

Another thing that can go wrong is error propagation. In a system with a
large number of equations (say, 100 or more), small rounding errors
early in the procedure can snowball to overwhelm the solution at the
end.

These issues, and many others like them, are outside the scope of this
book, but remember that just because Gauss\' method always works in
theory and just because a program correctly implements that method and
just because the answer appears on green-bar paper, doesn\'t mean that
answer is right. In practice, always use a package where experts have


# Linear Algebra/Linear Geometry of n-Space

|current=Linear Geometry of n-Space
|previous=Automation
who have seen the elements of vectors before, in calculus or physics,
this section is an optional review. However, later work will refer to
this material so it is not optional if it is not a review.*

In the first section, we had to do a bit of work to show that there are
only three types of solution sets--- singleton, empty, and infinite. But
in the special case of systems with two equations and two unknowns this
is easy to see. Draw each two-unknowns equation as a line in the plane
and then the two lines could have a unique intersection, be parallel, or
be the same line.

```{=html}
<TABLE border=0px cellpadding=20px style="text-align:center;margin:auto;">
```
```{=html}
<TR>
```
```{=html}
<TD>
```
*Unique solution*

```{=html}
<TD>
```
*No solutions*

```{=html}
<TD>
```
*Infinitely many*\
*solutions*

```{=html}
<TR>
```
```{=html}
<TD>
```
![](Linalg_unique_solution.png "Linalg_unique_solution.png"){width="150"}\
$\begin{array}{*{2}{rc}r}
\scriptstyle 3x  &\scriptstyle +  &\scriptstyle 2y  &\scriptstyle =  &\scriptstyle 7   \\[-5pt]
\scriptstyle x   &\scriptstyle -  &\scriptstyle y   &\scriptstyle =  &\scriptstyle -1
\end{array}$

```{=html}
<TD>
```
![](Linalg_no_solutions.png "Linalg_no_solutions.png"){width="150"}\
$\begin{array}{*{2}{rc}r}
\scriptstyle 3x  &\scriptstyle +  &\scriptstyle 2y  &\scriptstyle =  &\scriptstyle 7   \\[-5pt]
\scriptstyle 3x  &\scriptstyle +  &\scriptstyle 2y  &\scriptstyle =  &\scriptstyle 4
\end{array}$

```{=html}
<TD>
```
![](Linalg_infinitely_many_solutions.png "Linalg_infinitely_many_solutions.png"){width="150"}\
$\scriptstyle \begin{array}{*{2}{rc}r}
\scriptstyle 3x  &\scriptstyle +  &\scriptstyle 2y  &\scriptstyle =  &\scriptstyle 7   \\[-5pt]
\scriptstyle 6x  &\scriptstyle +  &\scriptstyle 4y  &\scriptstyle =  &\scriptstyle 14
\end{array}$

```{=html}
</TABLE>
```
These pictures don\'t prove the results from the prior section, which
apply to any number of linear equations and any number of unknowns, but
nonetheless they do help us to understand those results. This section
develops the ideas that we need to express our results from the prior
section, and from some future sections, geometrically. In particular,
while the two-dimensional case is familiar, to extend to systems with
more than two unknowns we shall need some higher-dimensional geometry.

|current=Linear Geometry of n-Space
|previous=Automation
|next=Vectors in Space}}`{=mediawiki}


# Linear Algebra/Vectors in Space

|current=Vectors in Space
|previous=Linear Geometry of n-Space

\"Higher-dimensional geometry\" sounds exotic. It is exotic---
interesting and eye-opening. But it isn\'t distant or unreachable.

We begin by defining one-dimensional space to be the set $\mathbb{R}^1$.
To see that definition is reasonable, draw a one-dimensional space

![](Linalg_line.png "Linalg_line.png"){width="800"}

and make the usual correspondence with $\mathbb{R}$: pick a point to
label $0$ and another to label $1$.

![](Linalg_line_with_unit.png "Linalg_line_with_unit.png"){width="800"}

Now, with a scale and a direction, finding the point corresponding to,
say $+2.17$, is easy--- start at $0$ and head in the direction of $1$
(i.e., the positive direction), but don\'t stop there, go $2.17$ times
as far.

The basic idea here, combining magnitude with direction, is the key to
extending to higher dimensions.

An object comprised of a magnitude and a direction is a **vector** (we
will use the same word as in the previous section because we shall show
below how to describe such an object with a column vector). We can draw
a vector as having some length, and pointing somewhere.

![](Linalg_vectors_1.png "Linalg_vectors_1.png")

There is a subtlety here--- these vectors

![](Linalg_vectors_2.png "Linalg_vectors_2.png"){width="150"}

are equal, even though they start in different places, because they have
equal lengths and equal directions. Again: those vectors are not just
alike, they are equal.

How can things that are in different places be equal? Think of a vector
as representing a displacement (\"vector\" is Latin for \"carrier\" or
\"traveler\"). These squares undergo the same displacement, despite that
those displacements start in different places.

![](Linalg_vectors_3.png "Linalg_vectors_3.png")

Sometimes, to emphasize this property vectors have of not being
anchored, they are referred to as **free** vectors. Thus, these free
vectors are equal as each is a displacement of one over and two up.

![](Linalg_vectors_4.png "Linalg_vectors_4.png")

More generally, vectors in the plane are the same if and only if they
have the same change in first components and the same change in second
components: the vector extending from $(a_1,a_2)$ to $(b_1,b_2)$ equals
the vector from $(c_1,c_2)$ to $(d_1,d_2)$ if and only if
$b_1-a_1=d_1-c_1$ and $b_2-a_2=d_2-c_2$.

An expression like \"the vector that, were it to start at $(a_1,a_2)$,
would extend to $(b_1,b_2)$\" is awkward. We instead describe such a
vector as

$$\begin{pmatrix} b_1-a_1 \\ b_2-a_2 \end{pmatrix}$$

so that, for instance, the \"one over and two up\" arrows shown above
picture this vector.

$$\begin{pmatrix} 1 \\ 2 \end{pmatrix}$$

We often draw the arrow as starting at the origin, and we then say it is
in the **canonical position** (or **natural position**). When the vector

$$\begin{pmatrix} b_1-a_1 \\ b_2-a_2 \end{pmatrix}$$

is in its canonical position then it extends to the endpoint
$(b_1-a_1,b_2-a_2)$.

We typically just refer to \"the point

$$\begin{pmatrix} 1 \\ 2 \end{pmatrix}$$\"

rather than \"the endpoint of the canonical position of\" that vector.

Thus, we will call both of these sets $\mathbb{R}^2$.

$$\{(x_1,x_2)\,\big|\, x_1,x_2\in\mathbb{R}\}
\qquad
\{\begin{pmatrix} x_1 \\ x_2 \end{pmatrix}\,\big|\, x_1,x_2\in\mathbb{R}\}$$

In the prior section we defined vectors and vector operations with an
algebraic motivation;

$$r\cdot\begin{pmatrix} v_1 \\ v_2 \end{pmatrix}
=
\begin{pmatrix} rv_1 \\ rv_2 \end{pmatrix}
\qquad
\begin{pmatrix} v_1 \\  v_2 \end{pmatrix}
+
\begin{pmatrix} w_1 \\ w_2 \end{pmatrix}
=
\begin{pmatrix} v_1+w_1 \\ v_2+w_2 \end{pmatrix}$$

we can now interpret those operations geometrically. For instance, if
$\vec{v}$ represents a displacement then $3\vec{v}\,$ represents a
displacement in the same direction but three times as far, and
$-1\vec{v}\,$ represents a displacement of the same distance as
$\vec{v}\,$ but in the opposite direction.

![](Linalg_scaled_vectors.png "Linalg_scaled_vectors.png")

And, where $\vec{v}$ and $\vec{w}$ represent displacements,
$\vec{v}+\vec{w}$ represents those displacements combined.

![](Linalg_vector_addition.png "Linalg_vector_addition.png")

The long arrow is the combined displacement in this sense: if, in one
minute, a ship\'s motion gives it the displacement relative to the earth
of $\vec{v}$ and a passenger\'s motion gives a displacement relative to
the ship\'s deck of $\vec{w}$, then $\vec{v}+\vec{w}$ is the
displacement of the passenger relative to the earth.

Another way to understand the vector sum is with the **parallelogram
rule**. Draw the parallelogram formed by the vectors
$\vec{v}_1,\vec{v}_2$ and then the sum $\vec{v}_1+\vec{v}_2$ extends
along the diagonal to the far corner.

![](Linalg_vector_addition_2.png "Linalg_vector_addition_2.png")

The above drawings show how vectors and vector operations behave in
$\mathbb{R}^2$. We can extend to $\mathbb{R}^3$, or to even
higher-dimensional spaces where we have no pictures, with the obvious
generalization: the free vector that, if it starts at
$(a_1,\ldots,a_n)$, ends at $(b_1,\ldots,b_n)$, is represented by this
column

$$\begin{pmatrix} b_1-a_1 \\ \vdots \\ b_n-a_n \end{pmatrix}$$

(vectors are equal if they have the same representation), we aren\'t too
careful to distinguish between a point and the vector whose canonical
representation ends at that point,

$$\mathbb{R}^n=
\{\begin{pmatrix} v_1 \\ \vdots \\ v_n \end{pmatrix}\,\big|\, v_1,\ldots,v_n\in\mathbb{R}\}$$

and addition and scalar multiplication are component-wise.

Having considered points, we now turn to the lines.

In $\mathbb{R}^2$, the line through $(1,2)$ and $(3,1)$ s comprised of
(the endpoints of) the vectors in this set

$$\{ \begin{pmatrix} 1 \\ 2 \end{pmatrix}+t\cdot\begin{pmatrix} 2 \\ -1 \end{pmatrix}\,\big|\, t\in\mathbb{R}\}$$

That description expresses this picture.

![](Linalg_vector_subtraction.png "Linalg_vector_subtraction.png")

The vector associated with the parameter $t$ has its whole body in the
line--- it is a **direction vector** for the line. Note that points on
the line to the left of $x=1$ are described using negative values of
$t$.

In $\mathbb{R}^3$, the line through $(1,2,1)$ and $(2,3,2)$ is the set
of (endpoints of) vectors of this form

![](Linalg_line_in_R3.png "Linalg_line_in_R3.png")

and lines in even higher-dimensional spaces work in the same way.

If a line uses one parameter, so that there is freedom to move back and
forth in one dimension, then a plane must involve two. For example, the
plane through the points $(1,0,5)$, $(2,1,-3)$, and $(-2,4,0.5)$
consists of (endpoints of) the vectors in

$$\{ \begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix}
+t\cdot\begin{pmatrix} 1 \\ 1 \\ -8 \end{pmatrix}
+s\cdot\begin{pmatrix} -3 \\ 4 \\ -4.5 \end{pmatrix}
\,\big|\, t,s\in\mathbb{R}      \}$$

(the column vectors associated with the parameters

$$\begin{pmatrix} 1 \\ 1 \\ -8 \end{pmatrix}
=
\begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}
-
\begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix}
\qquad
\begin{pmatrix} -3 \\ 4 \\ -4.5 \end{pmatrix}
=
\begin{pmatrix} -2 \\ 4 \\ 0.5 \end{pmatrix}
-
\begin{pmatrix} 1 \\ 0 \\ 5 \end{pmatrix}$$

are two vectors whose whole bodies lie in the plane). As with the line,
note that some points in this plane are described with negative $t$\'s
or negative $s$\'s or both.

A description of planes that is often encountered in algebra and
calculus uses a single equation as the condition that describes the
relationship among the first, second, and third coordinates of points in
a plane.

![](Linalg_plane.png "Linalg_plane.png")

The translation from such a description to the vector description that
we favor in this book is to think of the condition as a one-equation
linear system and parametrize $x=(1/2)(4-y-z)$.

![](Linalg_plane_with_basis.png "Linalg_plane_with_basis.png")

Generalizing from lines and planes, we define a **$k$-dimensional linear
surface** (or **$k$-flat**) in $\mathbb{R}^n$ to be
$\{\vec{p}+t_1\vec{v}_1+t_2\vec{v}_2+\cdots+t_k\vec{v}_k \,\big|\, t_1,\ldots ,t_k\in\mathbb{R}\}$
where $\vec{v}_1,\ldots,\vec{v}_k\in\mathbb{R}^n$. For example, in
$\mathbb{R}^4$,

$$\{\begin{pmatrix} 2 \\ \pi \\ 3 \\ -0.5 \end{pmatrix}
+t\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}
\,\big|\, t\in\mathbb{R}\}$$

is a line,

$$\{
\begin{pmatrix} 0 \\ 0 \\ 0 \\ 0 \end{pmatrix}
+t\begin{pmatrix} 1 \\ 1 \\ 0 \\ -1 \end{pmatrix}
+s\begin{pmatrix} 2 \\ 0 \\ 1 \\ 0 \end{pmatrix}
\,\big|\, t,s\in\mathbb{R}\}$$

is a plane, and

$$\{
\begin{pmatrix} 3 \\ 1 \\ -2 \\ 0.5 \end{pmatrix}
+r\begin{pmatrix} 0 \\ 0 \\ 0 \\ -1 \end{pmatrix}
+s\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix}
+t\begin{pmatrix} 2 \\ 0 \\ 1 \\ 0 \end{pmatrix}
\,\big|\, r,s,t\in\mathbb{R}\}$$

is a three-dimensional linear surface. Again, the intuition is that a
line permits motion in one direction, a plane permits motion in
combinations of two directions, etc.

A linear surface description can be misleading about the dimension---
this

$$L=\{
\begin{pmatrix} 1 \\ 0 \\ -1 \\ -2 \end{pmatrix}
+t\begin{pmatrix} 1 \\ 1 \\ 0 \\ -1 \end{pmatrix}
+s\begin{pmatrix} 2 \\ 2 \\ 0 \\ -2 \end{pmatrix}
\,\big|\, t,s\in\mathbb{R}\}$$

is a **degenerate** plane because it is actually a line.

$$L=\{
\begin{pmatrix} 1 \\ 0 \\ -1 \\ -2 \end{pmatrix}
+r\begin{pmatrix} 1 \\ 1 \\ 0 \\ -1 \end{pmatrix}
\,\big|\, r\in\mathbb{R}\}$$

We shall see in the Linear Independence section of Chapter Two what
relationships among vectors causes the linear surface they generate to
be degenerate.

We finish this subsection by restating our conclusions from the first
section in geometric terms. First, the solution set of a linear system
with $n$ unknowns is a linear surface in $\mathbb{R}^n$. Specifically,
it is a $k$-dimensional linear surface, where $k$ is the number of free
variables in an echelon form version of the system. Second, the solution
set of a homogeneous linear system is a linear surface passing through
the origin. Finally, we can view the general solution set of any linear
system as being the solution set of its associated homogeneous system
offset from the origin by a vector, namely by any particular solution.

## Exercises

## References

-       .

-       .


# Linear Algebra/Length and Angle Measures

|current=Length and Angle Measures
|previous=Vectors in Space
translated the first section\'s results about solution sets into
geometric terms for insight into how those sets look. But we must watch
out not to be mislead by our own terms; labeling subsets of
$\mathbb{R}^k$ of the forms
$\{\vec{p}+t\vec{v}\,\big|\, t\in\mathbb{R}\}$ and
$\{\vec{p}+t\vec{v}+s\vec{w}\,\big|\, t,s\in\mathbb{R}\}$ as \"lines\"
and \"planes\" doesn\'t make them act like the lines and planes of our
prior experience. Rather, we must ensure that the names suit the sets.
While we can\'t prove that the sets satisfy our intuition--- we can\'t
prove anything about intuition--- in this subsection we\'ll observe that
a result familiar from $\mathbb{R}^2$ and $\mathbb{R}^3$, when
generalized to arbitrary $\mathbb{R}^k$, supports the idea that a line
is straight and a plane is flat. Specifically, we\'ll see how to do
Euclidean geometry in a \"plane\" by giving a definition of the angle
between two $\mathbb{R}^n$ vectors in the plane that they generate.

We can use that definition to derive a formula for the angle between two
vectors. For a model of what to do, consider two vectors in
$\mathbb{R}^3$.

```{=html}
<center>
```
![](Linalg_two_vectors_in_R3.png "Linalg_two_vectors_in_R3.png"){width=""
height="200"}

```{=html}
</center>
```
Put them in canonical position and, in the plane that they determine,
consider the triangle formed by $\vec{u}$, $\vec{v}$, and
$\vec{u}-\vec{v}$.

```{=html}
<center>
```
![](Linalg_triangle_formed_by_two_vectors.png "Linalg_triangle_formed_by_two_vectors.png"){width=""
height="200"}

```{=html}
</center>
```
Apply the Law of Cosines, $|\vec{u}-\vec{v}\,|^2
=
|\vec{u}\,|^2+|\vec{v}\,|^2-
2\,|\vec{u}\,|\,|\vec{v}\,|\cos\theta$, where $\theta$ is the angle
between the vectors. Expand both sides

$$(u_1-v_1)^2+(u_2-v_2)^2+(u_3-v_3)^2$$

$$=(u_1^2+u_2^2+u_3^2)+(v_1^2+v_2^2+v_3^2)-
2\,|\vec{u}\,|\,|\vec{v}\,|\cos\theta$$

and simplify.

$$\theta
=
\arccos(\,\frac{u_1v_1+u_2v_2+u_3v_3}{
|\vec{u}\,|\,|\vec{v}\,| }\,)$$

In higher dimensions no picture suffices but we can make the same
argument analytically. First, the form of the numerator is clear--- it
comes from the middle terms of the squares $(u_1-v_1)^2$, $(u_2-v_2)^2$,
etc.

Note that the dot product of two vectors is a real number, not a vector,
and that the dot product of a vector from $\mathbb{R}^n$ with a vector
from $\mathbb{R}^m$ is defined only when $n$ equals $m$. Note also this
relationship between dot product and length: dotting a vector with
itself gives its length squared
$\vec{u}\cdot\vec{u}=u_1u_1+\cdots+u_nu_n=|\vec{u}\,|^2$.

Still reasoning with letters, but guided by the pictures, we use the
next theorem to argue that the triangle formed by $\vec{u}$, $\vec{v}$,
and $\vec{u}-\vec{v}$ in $\mathbb{R}^n$ lies in the planar subset of
$\mathbb{R}^n$ generated by $\vec{u}$ and $\vec{v}$.

This inequality is the source of the familiar saying, \"The shortest
distance between two points is in a straight line.\"

```{=html}
<center>
```
![](Linalg_triangle_inequality.png "Linalg_triangle_inequality.png"){width=""
height="150"}

```{=html}
</center>
```
This result supports the intuition that even in higher-dimensional
spaces, lines are straight and planes are flat. For any two points in a
linear surface, the line segment connecting them is contained in that
surface (this is easily checked from the definition). But if the surface
has a bend then that would allow for a shortcut (shown here grayed,
while the segment from $P$ to $Q$ that is contained in the surface is
solid).

```{=html}
<center>
```
![](Linalg_shortest_path_on_surface.png "Linalg_shortest_path_on_surface.png"){width=""
height="150"}

```{=html}
</center>
```
Because the Triangle Inequality says that in any $\mathbb{R}^n$, the
shortest cut between two endpoints is simply the line segment connecting
them, linear surfaces have no such bends.

Back to the definition of angle measure. The heart of the Triangle
Inequality\'s proof is the
\"$\vec{u}\cdot\vec{v}\leq |\vec{u}\,|\,|\vec{v}\,|$\" line. At first
glance, a reader might wonder if some pairs of vectors satisfy the
inequality in this way: while $\vec{u}\cdot\vec{v}$ is a large number,
with absolute value bigger than the right-hand side, it is a negative
large number. The next result says that no such pair of vectors exists.

The Cauchy-Schwarz inequality assures us that the next definition makes
sense because the fraction has absolute value less than or equal to one.

{ \|\\vec{u}\\,\|\\,\|\\vec{v}\\,\| }\\,) `</math>`{=html}

(the angle between the zero vector and any other vector is defined to be
a right angle). }}

Thus vectors from $\mathbb{R}^n$ are orthogonal (or perpendicular) if
and only if their dot product is zero.

) =\\arccos(\\frac{3}{\\sqrt{2}\\sqrt{13}}) `</math>`{=html}

approximately $0.94 \text{radians}$. Notice that these vectors are not
orthogonal. Although the $yz$-plane may appear to be perpendicular to
the $xy$-plane, in fact the two planes are that way only in the weak
sense that there are vectors in each orthogonal to all vectors in the
other. Not every vector in each is orthogonal to all vectors in the
other. }}

## Exercises

{\|\\vec{u}\\,\| }+\\frac{\\vec{v}}{\|\\vec{v}\\,\| } `</math>`{=html}

bisects the angle between them. Illustrate in $\mathbb{R}^2$. }}
`{{TextBox|1=
;Problem 27:
Verify that the definition of angle is dimensionally correct:
(1) if <math> k>0 </math> then the cosine of the angle between <math> k\vec{u} </math>
and <math> \vec{v} </math> equals the cosine of the angle between <math> \vec{u} </math>
and <math> \vec{v} </math>, and (2) if <math> k<0 </math> then the cosine of the angle
between <math> k\vec{u} </math> and <math> \vec{v} </math> is the negative of the cosine
of the angle between <math> \vec{u} </math> and <math> \vec{v} </math>.
}}`{=mediawiki}
`{{TextBox|1=
;Problem 28:
Show that the inner product operation is '''linear''': for
<math> \vec{u},\vec{v},\vec{w}\in\mathbb{R}^n </math> and <math> k,m\in\mathbb{R} </math>,
<math>\vec{u}\cdot(k\vec{v}+m\vec{w})=
k(\vec{u}\cdot\vec{v})+m(\vec{u}\cdot\vec{w})</math>.
}}`{=mediawiki}
`{{TextBox|1=
;Problem 29{{anchor|arithmetic geometric mean inequality}}:
The '''geometric mean'''
of two positive reals <math> x, y </math> is <math> \sqrt{xy} </math>.
It is analogous to the '''arithmetic mean'''
<math> (x+y)/2 </math>.
Use the Cauchy-Schwarz inequality to show that
the geometric mean of any <math> x,y\in\mathbb{R} </math> is less
than or equal to the arithmetic mean.
}}`{=mediawiki} `{{TextBox|1=
;? Problem 30:
A ship is sailing with speed and direction <math> \vec{v}_1 </math>; the wind blows
apparently (judging by the vane on the mast) in the direction of a vector
<math> \vec{a} </math>; on changing the direction and speed of the ship from
<math> \vec{v}_1 </math> to <math> \vec{v}_2 </math> the apparent wind is in the direction
of a vector <math> \vec{b} </math>.

Find the vector velocity of the wind {{harv|Ivanoff|Esty|1933}}.
}}`{=mediawiki} `{{TextBox|1=
;Problem 31:
Verify the Cauchy-Schwarz inequality by first proving
Lagrange's identity:

:<math>
\left(\sum_{1\leq j\leq n} a_jb_j \right)^2
=
\left(\sum_{1\leq j\leq n}a_j^2\right)
\left(\sum_{1\leq j\leq n}b_j^2\right)
-
\sum_{1\leq k < j\leq n}(a_kb_j-a_jb_k)^2
</math>

and then noting that the final term is positive.
(Recall the meaning

:<math>
\sum_{1\leq j\leq n}a_jb_j=
a_1b_1+a_2b_2+\cdots+a_nb_n
</math>

and

:<math>
\sum_{1\leq j\leq n}{a_j}^2=
{a_1}^2+{a_2}^2+\cdots+{a_n}^2
</math>

of the <math> \Sigma </math> notation.)
This result
is an improvement over Cauchy-Schwarz because it gives a formula for
the difference between the two sides.
Interpret that difference in <math> \mathbb{R}^2 </math>.
/Solutions/

## References

-
-
-


# Linear Algebra/Reduced Echelon Form

|current=Reduced Echelon Form
|previous=Length and Angle Measures
developing the mechanics of Gauss\' method, we observed that it can be
done in more than one way. One example is that we sometimes have to swap
rows and there can be more than one row to choose from. Another example
is that from this matrix

$$\begin{pmatrix}
2 &2 \\
4 &3
\end{pmatrix}$$

Gauss\' method could derive any of these echelon form matrices.

$$\begin{pmatrix}
2 &2 \\
0 &-1
\end{pmatrix}
\qquad
\begin{pmatrix}
1 &1 \\
0 &-1
\end{pmatrix}
\qquad
\begin{pmatrix}
2 &0 \\
0 &-1
\end{pmatrix}$$

The first results from $-2\rho_1+\rho_2$. The second comes from
following $(1/2)\rho_1$ with $-4\rho_1+\rho_2$. The third comes from
$-2\rho_1+\rho_2$ followed by $2\rho_2+\rho_1$ (after the first pivot
the matrix is already in echelon form so the second one is extra work
but it is nonetheless a legal row operation).

The fact that the echelon form outcome of Gauss\' method is not unique
leaves us with some questions. Will any two echelon form versions of a
system have the same number of free variables? Will they in fact have
exactly the same variables free? In this section we will answer both
questions \"yes\". We will do more than answer the questions. We will
give a way to decide if one linear system can be derived from another by
row operations. The answers to the two questions will follow from this
|current=Reduced Echelon Form
|previous=Length and Angle Measures
|next=Gauss-Jordan Reduction}}`{=mediawiki}