# Linear Algebra/Definition and Examples of Vector Spaces/Solutions
## Solutions
### Problem 1

Name the zero vector for each of these vector spaces.
* The space of degree three polynomials under the natural operations
* The space of $2\!\times \!4$ matrices
* The space $\{f:[0,1]\to \mathbb {R} \,{\big |}\,f{\text{ is continuous}}\}$
* The space of real-valued functions of one natural number variable

### Answer

* $0+0x+0x^{2}+0x^{3}$
* ${\begin{pmatrix}0&0&0&0\\0&0&0&0\end{pmatrix}}$
* The constant function $f(x)=0$
* The constant function $f(n)=0$
    This exercise is recommended for all readers.

### Problem 2

Find the additive inverse, in the vector space, of the vector.
* In ${\mathcal {P}}_{3}$, the vector $-3-2x+x^{2}$.
* In the space $2\!\times \!2$,
    ${\begin{pmatrix}1&-1\\0&3\end{pmatrix}}.$
* In $\{ae^{x}+be^{-x}\,{\big |}\,a,b\in \mathbb {R} \}$, the space of functions of the real variable $x$ under the natural operations, the vector $3e^{x}-2e^{-x}$.

### Answer

* $3+2x-x^{2}$
* ${\begin{pmatrix}-1&+1\\0&-3\end{pmatrix}}$
* $-3e^{x}+2e^{-x}$
    This exercise is recommended for all readers.

### Problem 3

Show that each of these is a vector space.
* The set of linear polynomials ${\mathcal {P}}_{1}=\{a_{0}+a_{1}x\,{\big |}\,a_{0},a_{1}\in \mathbb {R} \}$ under the usual polynomial addition and scalar multiplication operations.
* The set of $2\!\times \!2$ matrices with real entries under the usual matrix operations.
* The set of three-component row vectors with their usual operations.
* The set under the operations inherited from $\mathbb {R} ^{4}$.
    $L=\{{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\in \mathbb {R} ^{4}\,{\big |}\,x+y-z+w=0\}$

### Answer

Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some comments.
* This is just like Example 1.3; the zero element is $0+0x$.
* The zero element of this space is the $2\!\times \!2$ matrix of zeroes.
* The zero element is the vector of zeroes.
* Closure of addition involves noting that the sum is in $L$ because $(x_{1}+x_{2})+(y_{1}+y_{2})-(z_{1}+z_{2})+(w_{1}+w_{2})=(x_{1}+y_{1}-z_{1}+w_{1})+(x_{2}+y_{2}-z_{2}+w_{2})=0+0$. Closure of scalar multiplication is similar. Note that the zero element, the vector of zeroes, is in $L$.
    ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\\w_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\\w_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\y_{1}+y_{2}\\z_{1}+z_{2}\\w_{1}+w_{2}\end{pmatrix}}$
    This exercise is recommended for all readers.

### Problem 4

Show that each of these is not a vector space. (Hint. Start by listing two members of each set.)
* Under the operations inherited from $\mathbb {R} ^{3}$, this set
    $\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\,{\big |}\,x+y+z=1\}$
* Under the operations inherited from $\mathbb {R} ^{3}$, this set
    $\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\,{\big |}\,x^{2}+y^{2}+z^{2}=1\}$
* Under the usual matrix operations,
    $\{{\begin{pmatrix}a&1\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}$
* Under the usual polynomial operations, where $\mathbb {R} ^{+}$ is the set of reals greater than zero
    $\{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0},a_{1},a_{2}\in \mathbb {R} ^{+}\}$
* Under the inherited operations,
    $\{{\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\,{\big |}\,x+3y=4{\text{ and }}2x-y=3{\text{ and }}6x+4y=10\}$

### Answer

In each item the set is called $Q$. For some items, there are other correct ways to show that $Q$ is not a vector space.
* It is not closed under addition; it fails to meet condition 1.
    ${\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}\not \in Q$
* It is not closed under addition.
    ${\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}\not \in Q$
* It is not closed under addition.
    ${\begin{pmatrix}0&1\\0&0\end{pmatrix}},\,{\begin{pmatrix}1&1\\0&0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1&2\\0&0\end{pmatrix}}\not \in Q$
* It is not closed under scalar multiplication.
    $1+1x+1x^{2}\in Q\qquad -1\cdot (1+1x+1x^{2})\not \in Q$
* It is empty, violating condition 4.

### Problem 5

Define addition and scalar multiplication operations to make the complex numbers a vector space over $\mathbb {R}$.

### Answer

The usual operations $(v_{0}+v_{1}i)+(w_{0}+w_{1}i)=(v_{0}+w_{0})+(v_{1}+w_{1})i$ and $r(v_{0}+v_{1}i)=(rv_{0})+(rv_{1})i$ suffice. The check is easy.
    This exercise is recommended for all readers.

### Problem 6

Is the set of rational numbers a vector space over $\mathbb {R}$ under the usual addition and scalar multiplication operations?

### Answer

No, it is not closed under scalar multiplication since, e.g., $\pi \cdot 1$ is not a rational number.

### Problem 7

Show that the set of linear combinations of the variables $x,y,z$ is a vector space under the natural addition and scalar multiplication operations.

### Answer

The natural operations are $(v_{1}x+v_{2}y+v_{3}z)+(w_{1}x+w_{2}y+w_{3}z)=(v_{1}+w_{1})x+(v_{2}+w_{2})y+(v_{3}+w_{3})z$ and $r\cdot (v_{1}x+v_{2}y+v_{3}z)=(rv_{1})x+(rv_{2})y+(rv_{3})z$. The check that this is a vector space is easy; use Example 1.3 as a guide.

### Problem 8

Prove that this is not a vector space: the set of two-tall column vectors with real entries subject to these operations.
    ${\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}-x_{2}\\y_{1}-y_{2}\end{pmatrix}}\qquad r\cdot {\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}rx\\ry\end{pmatrix}}$

### Answer

The "$+$" operation is not commutative (that is, condition 2 is not met); producing two members of the set witnessing this assertion is easy.

### Problem 9

Prove or disprove that $\mathbb {R} ^{3}$ is a vector space under these operations.
* ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\quad {\text{and}}\quad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rx\\ry\\rz\end{pmatrix}}$
* ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\quad {\text{and}}\quad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}$

### Answer

* It is not a vector space.
    $(1+1)\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\neq {\begin{pmatrix}1\\0\\0\end{pmatrix}}+{\begin{pmatrix}1\\0\\0\end{pmatrix}}$
* It is not a vector space.
    $1\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\neq {\begin{pmatrix}1\\0\\0\end{pmatrix}}$
    This exercise is recommended for all readers.

### Problem 10

For each, decide if it is a vector space; the intended operations are the natural ones.
* The diagonal $2\!\times \!2$ matrices
    $\{{\begin{pmatrix}a&0\\0&b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}$
* This set of $2\!\times \!2$ matrices
    $\{{\begin{pmatrix}x&x+y\\x+y&y\end{pmatrix}}\,{\big |}\,x,y\in \mathbb {R} \}$
* This set
    $\{{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\in \mathbb {R} ^{4}\,{\big |}\,x+y+z+w=1\}$
* The set of functions $\{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,df/dx+2f=0\}$
* The set of functions $\{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,df/dx+2f=1\}$

### Answer

For each "yes" answer, you must give a check of all the conditions given in the definition of a vector space. For each "no" answer, give a specific example of the failure of one of the conditions.
* Yes.
* Yes.
* No, it is not closed under addition. The vector of all $1/4$'s, when added to itself, makes a nonmember.
* Yes.
* No, $f(x)=e^{-2x}+(1/2)$ is in the set but $2\cdot f$ is not (that is, condition 6 fails).
    This exercise is recommended for all readers.

### Problem 11

Prove or disprove that this is a vector space: the real-valued functions $f$ of one real variable such that $f(7)=0$.

### Answer

It is a vector space. Most conditions of the definition of vector space are routine; we here check only closure. For addition, $(f_{1}+f_{2})\,(7)=f_{1}(7)+f_{2}(7)=0+0=0$. For scalar multiplication, $(r\cdot f)\,(7)=rf(7)=r0=0$.
    This exercise is recommended for all readers.

### Problem 12

Show that the set $\mathbb {R} ^{+}$ of positive reals is a vector space when "$x+y$" is interpreted to mean the product of $x$ and $y$ (so that $2+3$ is $6$), and "$r\cdot x$" is interpreted as the $r$-th power of $x$.

### Answer

We check Definition 1.1.
First, closure under "$+$" holds because the product of two positive reals is a positive real. The second condition is satisfied because real multiplication commutes. Similarly, as real multiplication associates, the third checks. For the fourth condition, observe that multiplying a number by $1\in \mathbb {R} ^{+}$ won't change the number. Fifth, any positive real has a reciprocal that is a positive real.
The sixth, closure under "$\cdot$", holds because any power of a positive real is a positive real. The seventh condition is just the rule that $v^{r+s}$ equals the product of $v^{r}$ and $v^{s}$. The eight condition says that $(vw)^{r}=v^{r}w^{r}$. The ninth condition asserts that $(v^{r})^{s}=v^{rs}$. The final condition says that $v^{1}=v$.

### Problem 13

Is $\{(x,y)\,{\big |}\,x,y\in \mathbb {R} \}$ a vector space under these operations?
* $(x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})$ and $r\cdot (x,y)=(rx,y)$
* $(x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})$ and $r\cdot (x,y)=(rx,0)$

### Answer

* No: $1\cdot (0,1)+1\cdot (0,1)\neq (1+1)\cdot (0,1)$.
* No; the same calculation as the prior answer shows a condition in the definition of a vector space that is violated. Another example of a violation of the conditions for a vector space is that $1\cdot (0,1)\neq (0,1)$.

### Problem 14

Prove or disprove that this is a vector space: the set of polynomials of degree greater than or equal to two, along with the zero polynomial.

### Answer

It is not a vector space since it is not closed under addition, as $(x^{2})+(1+x-x^{2})$ is not in the set.

### Problem 15

At this point "the same" is only an intuition, but nonetheless for each vector space identify the $k$ for which the space is "the same" as $\mathbb {R} ^{k}$.
* The $2\!\times \!3$ matrices under the usual operations
* The $n\!\times \!m$ matrices (under their usual operations)
* This set of $2\!\times \!2$ matrices
    $\{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}$
* This set of $2\!\times \!2$ matrices
    $\{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a+b+c=0\}$

### Answer

* $6$
* $nm$
* $3$
* To see that the answer is $2$, rewrite it as so that there are two parameters.
    $\{{\begin{pmatrix}a&0\\b&-a-b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}$
    This exercise is recommended for all readers.

### Problem 16

Using ${\vec {+}}$ to represent vector addition and $\,{\vec {\cdot }}\,$ for scalar multiplication, restate the definition of vector space.

### Answer

A vector space (over $\mathbb {R}$) consists of a set $V$ along with two operations "${\mathbin {\vec {+}}}$" and "${\mathbin {\vec {\cdot }}}$" subject to these conditions. Where ${\vec {v}},{\vec {w}}\in V$,
* their vector sum ${\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}$ is an element of $V$. If ${\vec {u}},{\vec {v}},{\vec {w}}\in V$ then
* ${\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}={\vec {w}}{\mathbin {\vec {+}}}{\vec {v}}$ and
* $({\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}){\mathbin {\vec {+}}}{\vec {u}}={\vec {v}}{\mathbin {\vec {+}}}({\vec {w}}{\mathbin {\vec {+}}}{\vec {u}})$.
* There is a zero vector ${\vec {0}}\in V$ such that ${\vec {v}}{\mathbin {\vec {+}}}{\vec {0}}={\vec {v}}\,$ for all ${\vec {v}}\in V$.
* Each ${\vec {v}}\in V$ has an additive inverse ${\vec {w}}\in V$ such that ${\vec {w}}{\mathbin {\vec {+}}}{\vec {v}}={\vec {0}}$. If $r,s$ are scalars, that is, members of $\mathbb {R}$), and ${\vec {v}},{\vec {w}}\in V$ then
* each scalar multiple $r\cdot {\vec {v}}$ is in $V$. If $r,s\in \mathbb {R}$ and ${\vec {v}},{\vec {w}}\in V$ then
* $(r+s)\cdot {\vec {v}}=r\cdot {\vec {v}}{\mathbin {\vec {+}}}s\cdot {\vec {v}}$, and
* $r{\mathbin {\vec {\cdot }}}({\vec {v}}+{\vec {w}})=r{\mathbin {\vec {\cdot }}}{\vec {v}}+r{\mathbin {\vec {\cdot }}}{\vec {w}}$, and
* $(rs){\mathbin {\vec {\cdot }}}{\vec {v}}=r{\mathbin {\vec {\cdot }}}(s{\mathbin {\vec {\cdot }}}{\vec {v}})$, and
* $1{\mathbin {\vec {\cdot }}}{\vec {v}}={\vec {v}}$.
    This exercise is recommended for all readers.

### Problem 17

Prove these.
* Any vector is the additive inverse of the additive inverse of itself.
* Vector addition left-cancels: if ${\vec {v}},{\vec {s}},{\vec {t}}\in V$ then ${\vec {v}}+{\vec {s}}={\vec {v}}+{\vec {t}}\,$ implies that ${\vec {s}}={\vec {t}}$.

### Answer

* Let $V$ be a vector space, assume that ${\vec {v}}\in V$, and assume that ${\vec {w}}\in V$ is the additive inverse of ${\vec {v}}$ so that ${\vec {w}}+{\vec {v}}={\vec {0}}$. Because addition is commutative, ${\vec {0}}={\vec {w}}+{\vec {v}}={\vec {v}}+{\vec {w}}$, so therefore ${\vec {v}}$ is also the additive inverse of ${\vec {w}}$.
* Let $V$ be a vector space and suppose ${\vec {v}},{\vec {s}},{\vec {t}}\in V$. The additive inverse of ${\vec {v}}$ is $-{\vec {v}}$ so ${\vec {v}}+{\vec {s}}={\vec {v}}+{\vec {t}}$ gives that $-{\vec {v}}+{\vec {v}}+{\vec {s}}=-{\vec {v}}+{\vec {v}}+{\vec {t}}$, which says that ${\vec {0}}+{\vec {s}}={\vec {0}}+{\vec {t}}$ and so ${\vec {s}}={\vec {t}}$.

### Problem 18

The definition of vector spaces does not explicitly say that ${\vec {0}}+{\vec {v}}={\vec {v}}$ (it instead says that ${\vec {v}}+{\vec {0}}={\vec {v}}$). Show that it must nonetheless hold in any vector space.

### Answer

Addition is commutative, so in any vector space, for any vector ${\vec {v}}$ we have that ${\vec {v}}={\vec {v}}+{\vec {0}}={\vec {0}}+{\vec {v}}$.
    This exercise is recommended for all readers.

### Problem 19

Prove or disprove that this is a vector space: the set of all matrices, under the usual operations.

### Answer

It is not a vector space since addition of two matrices of unequal sizes is not defined, and thus the set fails to satisfy the closure condition.

### Problem 20

In a vector space every element has an additive inverse. Can some elements have two or more?

### Answer

Each element of a vector space has one and only one additive inverse.
For, let $V$ be a vector space and suppose that ${\vec {v}}\in V$. If ${\vec {w}}_{1},{\vec {w}}_{2}\in V$ are both additive inverses of ${\vec {v}}$ then consider ${\vec {w}}_{1}+{\vec {v}}+{\vec {w}}_{2}$. On the one hand, we have that it equals ${\vec {w}}_{1}+({\vec {v}}+{\vec {w}}_{2})={\vec {w}}_{1}+{\vec {0}}={\vec {w}}_{1}$. On the other hand we have that it equals $({\vec {w}}_{1}+{\vec {v}})+{\vec {w}}_{2}={\vec {0}}+{\vec {w}}_{2}={\vec {w}}_{2}$. Therefore, ${\vec {w}}_{1}={\vec {w}}_{2}$.

### Problem 21

* Prove that every point, line, or plane thru the origin in $\mathbb {R} ^{3}$ is a vector space under the inherited operations.
* What if it doesn't contain the origin?

### Answer

* Every such set has the form $\{r\cdot {\vec {v}}+s\cdot {\vec {w}}\,{\big |}\,r,s\in \mathbb {R} \}$ where either or both of ${\vec {v}},{\vec {w}}$ may be ${\vec {0}}$. With the inherited operations, closure of addition $(r_{1}{\vec {v}}+s_{1}{\vec {w}})+(r_{2}{\vec {v}}+s_{2}{\vec {w}})=(r_{1}+r_{2}){\vec {v}}+(s_{1}+s_{2}){\vec {w}}$ and scalar multiplication $c(r{\vec {v}}+s{\vec {w}})=(cr){\vec {v}}+(cs){\vec {w}}$ are easy. The other conditions are also routine.
* No such set can be a vector space under the inherited operations because it does not have a zero element.
    This exercise is recommended for all readers.

### Problem 22

Using the idea of a vector space we can easily reprove that the solution set of a homogeneous linear system has either one element or infinitely many elements. Assume that ${\vec {v}}\in V$ is not ${\vec {0}}$.
* Prove that $r\cdot {\vec {v}}={\vec {0}}$ if and only if $r=0$.
* Prove that $r_{1}\cdot {\vec {v}}=r_{2}\cdot {\vec {v}}$ if and only if $r_{1}=r_{2}$.
* Prove that any nontrivial vector space is infinite.
* Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion.

### Answer

Assume that ${\vec {v}}\in V$ is not ${\vec {0}}$.
* One direction of the if and only if is clear: if $r=0$ then $r\cdot {\vec {v}}={\vec {0}}$. For the other way, let $r$ be a nonzero scalar. If $r{\vec {v}}={\vec {0}}$ then $1\cdot {\vec {v}}=(r/r)\cdot {\vec {v}}=(1/r)\cdot r{\vec {v}}=(1/r)\cdot {\vec {0}}$ shows that ${\vec {v}}={\vec {0}}$, contrary to the assumption.
* Where $r_{1},r_{2}$ are scalars, $r_{1}{\vec {v}}=r_{2}{\vec {v}}\,$ holds if and only if $(r_{1}-r_{2}){\vec {v}}={\vec {0}}$. By the prior item, then $r_{1}-r_{2}=0$.
* A nontrivial space has a vector ${\vec {v}}\neq {\vec {0}}$. Consider the set $\{k\cdot {\vec {v}}\,{\big |}\,k\in \mathbb {R} \}$. By the prior item this set is infinite.
* The solution set is either trivial, or nontrivial. In the second case, it is infinite.

### Problem 23

Is this a vector space under the natural operations: the real-valued functions of one real variable that are differentiable?

### Answer

Yes. A theorem of first semester calculus says that a sum of differentiable functions is differentiable and that $(f+g)^{\prime }=f^{\prime }+g^{\prime }$, and that a multiple of a differentiable function is differentiable and that $(r\cdot f)^{\prime }=r\,f^{\prime }$.

### Problem 24

A vector space over the complex numbers $\mathbb {C}$ has the same definition as a vector space over the reals except that scalars are drawn from $\mathbb {C}$ instead of from $\mathbb {R}$. Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply: $(a_{0}+a_{1}i)+(b_{0}+b_{1}i)=(a_{0}+b_{0})+(a_{1}+b_{1})i$ and $(a_{0}+a_{1}i)(b_{0}+b_{1}i)=(a_{0}b_{0}-a_{1}b_{1})+(a_{0}b_{1}+a_{1}b_{0})i$.)
* The set of degree two polynomials with complex coefficients
* This set
    $\{{\begin{pmatrix}0&a\\b&0\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {C} {\text{ and }}a+b=0+0i\}$

### Answer

The check is routine. Note that "1" is $1+0i$ and the zero elements are these.
* $(0+0i)+(0+0i)x+(0+0i)x^{2}$
* ${\begin{pmatrix}0+0i&0+0i\\0+0i&0+0i\end{pmatrix}}$

### Problem 25

Name a property shared by all of the $\mathbb {R} ^{n}$'s but not listed as a requirement for a vector space.

### Answer

Notably absent from the definition of a vector space is a distance measure.
    This exercise is recommended for all readers.

### Problem 26

* Prove that a sum of four vectors ${\vec {v}}_{1},\ldots ,{\vec {v}}_{4}\in V$ can be associated in any way without changing the result. This allows us to simply write "${\vec {v}}_{1}+{\vec {v}}_{2}+{\vec {v}}_{3}+{\vec {v}}_{4}$" without ambiguity.
    ${\begin{array}{rl}(({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+{\vec {v}}_{4}&=({\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3}))+{\vec {v}}_{4}\\&=({\vec {v}}_{1}+{\vec {v}}_{2})+({\vec {v}}_{3}+{\vec {v}}_{4})\\&={\vec {v}}_{1}+(({\vec {v}}_{2}+{\vec {v}}_{3})+{\vec {v}}_{4})\\&={\vec {v}}_{1}+({\vec {v}}_{2}+({\vec {v}}_{3}+{\vec {v}}_{4}))\end{array}}$
* Prove that any two ways of associating a sum of any number of vectors give the same sum. (Hint. Use induction on the number of vectors.)

### Answer

* A small rearrangement does the trick. Each equality above follows from the associativity of three vectors that is given as a condition in the definition of a vector space. For instance, the second "$=$" applies the rule $({\vec {w}}_{1}+{\vec {w}}_{2})+{\vec {w}}_{3}={\vec {w}}_{1}+({\vec {w}}_{2}+{\vec {w}}_{3})$ by taking ${\vec {w}}_{1}$ to be ${\vec {v}}_{1}+{\vec {v}}_{2}$, taking ${\vec {w}}_{2}$ to be ${\vec {v}}_{3}$, and taking ${\vec {w}}_{3}$ to be ${\vec {v}}_{4}$.
    ${\begin{array}{rl}({\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3}))+{\vec {v}}_{4}&=(({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+{\vec {v}}_{4}\\&=({\vec {v}}_{1}+{\vec {v}}_{2})+({\vec {v}}_{3}+{\vec {v}}_{4})\\&={\vec {v}}_{1}+({\vec {v}}_{2}+({\vec {v}}_{3}+{\vec {v}}_{4}))\\&={\vec {v}}_{1}+(({\vec {v}}_{2}+{\vec {v}}_{3})+{\vec {v}}_{4})\end{array}}$
* The base case for induction is the three vector case. This case ${\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3})=({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3}$ is required of any triple of vectors by the definition of a vector space. For the inductive step, assume that any two sums of three vectors, any two sums of four vectors, ..., any two sums of $k$ vectors are equal no matter how the sums are parenthesized. We will show that any sum of $k+1$ vectors equals this one $((\cdots (({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+\cdots )+{\vec {v}}_{k})+{\vec {v}}_{k+1}$. Any parenthesized sum has an outermost "$+$". Assume that it lies between ${\vec {v}}_{m}$ and ${\vec {v}}_{m+1}$ so the sum looks like this. The second half involves fewer than $k+1$ additions, so by the inductive hypothesis we can re-parenthesize it so that it reads left to right from the inside out, and in particular, so that its outermost "$+$" occurs right before ${\vec {v}}_{k+1}$. Apply the associativity of the sum of three things and finish by applying the inductive hypothesis inside these outermost parenthesis.
    $(\cdots \,({\vec {v}}_{1}+(\cdots +{\vec {v}}_{m}))\,\cdots )+(\cdots \,({\vec {v}}_{m+1}+(\cdots +{\vec {v}}_{k+1}))\,\cdots )$
    $=(\cdots \,({\vec {v}}_{1}+(\,\cdots \,+{\vec {v}}_{m}))\,\cdots )+(\cdots ({\vec {v}}_{m+1}+(\cdots +{\vec {v}}_{k})\cdots )+{\vec {v}}_{k+1})$
    $=((\,\cdots \,({\vec {v}}_{1}+(\cdots +{\vec {v}}_{m})\,\cdots \,)+(\,\cdots \,({\vec {v}}_{m+1}+(\cdots \,{\vec {v}}_{k}))\cdots )+{\vec {v}}_{k+1}$

### Problem 27

For any vector space, a subset that is itself a vector space under the inherited operations (e.g., a plane through the origin inside of $\mathbb {R} ^{3}$) is a subspace.
* Show that $\{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0}+a_{1}+a_{2}=0\}$ is a subspace of the vector space of degree two polynomials.
* Show that this is a subspace of the $2\!\times \!2$ matrices.
    $\{{\begin{pmatrix}a&b\\c&0\end{pmatrix}}\,{\big |}\,a+b=0\}$
* Show that a nonempty subset $S$ of a real vector space is a subspace if and only if it is closed under linear combinations of pairs of vectors: whenever $c_{1},c_{2}\in \mathbb {R}$ and ${\vec {s}}_{1},{\vec {s}}_{2}\in S$ then the combination $c_{1}{\vec {v}}_{1}+c_{2}{\vec {v}}_{2}$ is in $S$.

### Answer

* We outline the check of the conditions from Definition 1.1. Additive closure holds because if $a_{0}+a_{1}+a_{2}=0$ and $b_{0}+b_{1}+b_{2}=0$ then is in the set since $(a_{0}+b_{0})+(a_{1}+b_{1})+(a_{2}+b_{2})=(a_{0}+a_{1}+a_{2})+(b_{0}+b_{1}+b_{2})$ is zero. The second through fifth conditions are easy. Closure under scalar multiplication holds because if $a_{0}+a_{1}+a_{2}=0$ then is in the set as $ra_{0}+ra_{1}+ra_{2}=r(a_{0}+a_{1}+a_{2})$ is zero. The remaining conditions here are also easy.
    $(a_{0}+a_{1}x+a_{2}x^{2})+(b_{0}+b_{1}x+b_{2}x^{2})=(a_{0}+b_{0})+(a_{1}+b_{1})x+(a_{2}+b_{2})x^{2}$
    $r\cdot (a_{0}+a_{1}x+a_{2}x^{2})=(ra_{0})+(ra_{1})x+(ra_{2})x^{2}$
* This is similar to the prior answer.
* Call the vector space $V$. We have two implications: left to right, if $S$ is a subspace then it is closed under linear combinations of pairs of vectors and, right to left, if a nonempty subset is closed under linear combinations of pairs of vectors then it is a subspace. The left to right implication is easy; we here sketch the other one by assuming $S$ is nonempty and closed, and checking the conditions of Definition 1.1. First, to show closure under addition, if ${\vec {s}}_{1},{\vec {s}}_{2}\in S$ then ${\vec {s}}_{1}+{\vec {s}}_{2}\in S$ as ${\vec {s}}_{1}+{\vec {s}}_{2}=1\cdot {\vec {s}}_{1}+1\cdot {\vec {s}}_{2}$. Second, for any ${\vec {s}}_{1},{\vec {s}}_{2}\in S$, because addition is inherited from $V$, the sum ${\vec {s}}_{1}+{\vec {s}}_{2}$ in $S$ equals the sum ${\vec {s}}_{1}+{\vec {s}}_{2}$ in $V$ and that equals the sum ${\vec {s}}_{2}+{\vec {s}}_{1}$ in $V$ and that in turn equals the sum ${\vec {s}}_{2}+{\vec {s}}_{1}$ in $S$. The argument for the third condition is similar to that for the second. For the fourth, suppose that ${\vec {s}}$ is in the nonempty set $S$ and note that $0\cdot {\vec {s}}={\vec {0}}\in S$; showing that the ${\vec {0}}$ of $V$ acts under the inherited operations as the additive identity of $S$ is easy. The fifth condition is satisfied because for any ${\vec {s}}\in S$ closure under linear combinations shows that the vector $0\cdot {\vec {0}}+(-1)\cdot {\vec {s}}$ is in $S$; showing that it is the additive inverse of ${\vec {s}}$ under the inherited operations is routine. The proofs for the remaining conditions are similar.