# Linear Algebra/Subspaces and Spanning sets/Solutions
## Solutions

    This exercise is recommended for all readers.

### Problem 1

Which of these subsets of the vector space of $2\!\times \!2$ matrices are subspaces under the inherited operations? For each one that is a subspace, parametrize its description. For each that is not, give a condition that fails.
* $\{{\begin{pmatrix}a&0\\0&b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}$
* $\{{\begin{pmatrix}a&0\\0&b\end{pmatrix}}\,{\big |}\,a+b=0\}$
* $\{{\begin{pmatrix}a&0\\0&b\end{pmatrix}}\,{\big |}\,a+b=5\}$
* $\{{\begin{pmatrix}a&c\\0&b\end{pmatrix}}\,{\big |}\,a+b=0,c\in \mathbb {R} \}$

### Answer

By Lemma 2.9, to see if each subset of ${\mathcal {M}}_{2\!\times \!2}$ is a subspace, we need only check if it is nonempty and closed.
* Yes, it is easily checked to be nonempty and closed. This is a parametrization. By the way, the parametrization also shows that it is a subspace, it is given as the span of the two-matrix set, and any span is a subspace.
    $\{a{\begin{pmatrix}1&0\\0&0\end{pmatrix}}+b{\begin{pmatrix}0&0\\0&1\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}$
* Yes; it is easily checked to be nonempty and closed. Alternatively, as mentioned in the prior answer, the existence of a parametrization shows that it is a subspace. For the parametrization, the condition $a+b=0$ can be rewritten as $a=-b$. Then we have this.
    $\{{\begin{pmatrix}-b&0\\0&b\end{pmatrix}}\,{\big |}\,b\in \mathbb {R} \}=\{b{\begin{pmatrix}-1&0\\0&1\end{pmatrix}}\,{\big |}\,b\in \mathbb {R} \}$
* No. It is not closed under addition. For instance, is not in the set. (This set is also not closed under scalar multiplication, for instance, it does not contain the zero matrix.)
    ${\begin{pmatrix}5&0\\0&0\end{pmatrix}}+{\begin{pmatrix}5&0\\0&0\end{pmatrix}}={\begin{pmatrix}10&0\\0&0\end{pmatrix}}$
* Yes.
    $\{b{\begin{pmatrix}-1&0\\0&1\end{pmatrix}}+c{\begin{pmatrix}0&1\\0&0\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}$
    This exercise is recommended for all readers.

### Problem 2

Is this a subspace of ${\mathcal {P}}_{2}$: $\{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0}+2a_{1}+a_{2}=4\}$? If it is then parametrize its description.

### Answer

No, it is not closed. In particular, it is not closed under scalar multiplication because it does not contain the zero polynomial.
    This exercise is recommended for all readers.

### Problem 3

Decide if the vector lies in the span of the set, inside of the space.
* ${\begin{pmatrix}2\\0\\1\end{pmatrix}}$, $\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\}$, in $\mathbb {R} ^{3}$
* $x-x^{3}$, $\{x^{2},2x+x^{2},x+x^{3}\}$, in ${\mathcal {P}}_{3}$
* ${\begin{pmatrix}0&1\\4&2\end{pmatrix}}$, $\{{\begin{pmatrix}1&0\\1&1\end{pmatrix}},{\begin{pmatrix}2&0\\2&3\end{pmatrix}}\}$, in ${\mathcal {M}}_{2\!\times \!2}$

### Answer

* Yes, solving the linear system arising from $r_{1}=2$ and $r_{2}=1$.
    $r_{1}{\begin{pmatrix}1\\0\\0\end{pmatrix}}+r_{2}{\begin{pmatrix}0\\0\\1\end{pmatrix}}={\begin{pmatrix}2\\0\\1\end{pmatrix}}$
* Yes; the linear system arising from $r_{1}(x^{2})+r_{2}(2x+x^{2})+r_{3}(x+x^{3})=x-x^{3}$ gives that $-1(x^{2})+1(2x+x^{2})-1(x+x^{3})=x-x^{3}$.
    ${\begin{array}{*{3}{rc}r}&&2r_{2}&+&r_{3}&=&1\\r_{1}&+&r_{2}&&&=&0\\&&&&r_{3}&=&-1\end{array}}$
* No; any combination of the two given matrices has a zero in the upper right.

### Problem 4

Which of these are members of the span $[\{\cos ^{2}x,\sin ^{2}x\}]$ in the vector space of real-valued functions of one real variable?
* $f(x)=1$
* $f(x)=3+x^{2}$
* $f(x)=\sin x$
* $f(x)=\cos(2x)$

### Answer

* Yes; it is in that span since $1\cdot \cos ^{2}x+1\cdot \sin ^{2}x=f(x)$.
* No, since $r_{1}\cos ^{2}x+r_{2}\sin ^{2}x=3+x^{2}$ has no scalar solutions that work for all $x$. For instance, setting $x$ to be $0$ and $\pi$ gives the two equations $r_{1}\cdot 1+r_{2}\cdot 0=3$ and $r_{1}\cdot 1+r_{2}\cdot 0=3+\pi ^{2}$, which are not consistent with each other.
* No; consider what happens on setting $x$ to be $\pi /2$ and $3\pi /2$.
* Yes, $\cos(2x)=1\cdot \cos ^{2}(x)-1\cdot \sin ^{2}(x)$.
    This exercise is recommended for all readers.

### Problem 5

Which of these sets spans $\mathbb {R} ^{3}$? That is, which of these sets has the property that any three-tall vector can be expressed as a suitable linear combination of the set's elements?
* $\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\2\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\3\end{pmatrix}}\}$
* $\{{\begin{pmatrix}2\\0\\1\end{pmatrix}},{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\}$
* $\{{\begin{pmatrix}1\\1\\0\end{pmatrix}},{\begin{pmatrix}3\\0\\0\end{pmatrix}}\}$
* $\{{\begin{pmatrix}1\\0\\1\end{pmatrix}},{\begin{pmatrix}3\\1\\0\end{pmatrix}},{\begin{pmatrix}-1\\0\\0\end{pmatrix}},{\begin{pmatrix}2\\1\\5\end{pmatrix}}\}$
* $\{{\begin{pmatrix}2\\1\\1\end{pmatrix}},{\begin{pmatrix}3\\0\\1\end{pmatrix}},{\begin{pmatrix}5\\1\\2\end{pmatrix}},{\begin{pmatrix}6\\0\\2\end{pmatrix}}\}$

### Answer

* Yes, for any $x,y,z\in \mathbb {R}$ this equation has the solution $r_{1}=x$, $r_{2}=y/2$, and $r_{3}=z/3$.
    $r_{1}{\begin{pmatrix}1\\0\\0\end{pmatrix}}+r_{2}{\begin{pmatrix}0\\2\\0\end{pmatrix}}+r_{3}{\begin{pmatrix}0\\0\\3\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
* Yes, the equation gives rise to this so that, given any $x$, $y$, and $z$, we can compute that $r_{3}=(-1/2)x+(1/2)y+z$, $r_{2}=y$, and $r_{1}=(1/2)x-(1/2)y$.
    $r_{1}{\begin{pmatrix}2\\0\\1\end{pmatrix}}+r_{2}{\begin{pmatrix}1\\1\\0\end{pmatrix}}+r_{3}{\begin{pmatrix}0\\0\\1\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    ${\begin{array}{*{3}{rc}r}2r_{1}&+&r_{2}&&&=&x\\&&r_{2}&&&=&y\\r_{1}&&&+&r_{3}&=&z\\\end{array}}\;{\xrightarrow[]{-(1/2)\rho _{1}+\rho _{3}}}\;\;{\xrightarrow[]{(1/2)\rho _{2}+\rho _{3}}}\;{\begin{array}{*{3}{rc}r}2r_{1}&+&r_{2}&&&=&x\\&&r_{2}&&&=&y\\&&&&r_{3}&=&-(1/2)x+(1/2)y+z\\\end{array}}$
* No. In particular, the vector cannot be gotten as a linear combination since the two given vectors both have a third component of zero.
    ${\begin{pmatrix}0\\0\\1\end{pmatrix}}$
* Yes. The equation leads to this reduction. We have infinitely many solutions. We can, for example, set $r_{4}$ to be zero and solve for $r_{3}$, $r_{2}$, and $r_{1}$ in terms of $x$, $y$, and $z$ by the usual methods of back-substitution.
    $r_{1}{\begin{pmatrix}1\\0\\1\end{pmatrix}}+r_{2}{\begin{pmatrix}3\\1\\0\end{pmatrix}}+r_{3}{\begin{pmatrix}-1\\0\\0\end{pmatrix}}+r_{4}{\begin{pmatrix}2\\1\\5\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    $\left({\begin{array}{*{4}{c}|c}1&3&-1&2&x\\0&1&0&1&y\\1&0&0&5&z\end{array}}\right){\xrightarrow[]{-\rho _{1}+\rho _{3}}}\;{\xrightarrow[]{3\rho _{2}+\rho _{3}}}\left({\begin{array}{*{4}{c}|c}1&3&-1&2&x\\0&1&0&1&y\\0&0&1&6&-x+3y+z\end{array}}\right)$
* No. The equation leads to this reduction. This shows that not every three-tall vector can be so expressed. Only the vectors satisfying the restriction that $-(1/3)x-(1/3)y+z=0$ are in the span. (To see that any such vector is indeed expressible, take $r_{3}$ and $r_{4}$ to be zero and solve for $r_{1}$ and $r_{2}$ in terms of $x$, $y$, and $z$ by back-substitution.)
    $r_{1}{\begin{pmatrix}2\\1\\1\end{pmatrix}}+r_{2}{\begin{pmatrix}3\\0\\1\end{pmatrix}}+r_{3}{\begin{pmatrix}5\\1\\2\end{pmatrix}}+r_{4}{\begin{pmatrix}6\\0\\2\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    $\left({\begin{array}{*{4}{c}|c}2&3&5&6&x\\1&0&1&0&y\\1&1&2&2&z\end{array}}\right){\xrightarrow[{-(1/2)\rho _{1}+\rho _{3}}]{-(1/2)\rho _{1}+\rho _{2}}}\;{\xrightarrow[]{-(1/3)\rho _{2}+\rho _{3}}}\left({\begin{array}{*{4}{c}|c}2&3&5&6&x\\0&-3/2&-3/2&-3&-(1/2)x+y\\0&0&0&0&-(1/3)x-(1/3)y+z\end{array}}\right)$
    This exercise is recommended for all readers.

### Problem 6

Parametrize each subspace's description. Then express each subspace as a span.
* The subset $\{{\begin{pmatrix}a&b&c\end{pmatrix}}\,{\big |}\,a-c=0\}$ of the three-wide row vectors
* This subset of ${\mathcal {M}}_{2\!\times \!2}$
    $\{{\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,{\big |}\,a+d=0\}$
* This subset of ${\mathcal {M}}_{2\!\times \!2}$
    $\{{\begin{pmatrix}a&b\\c&d\end{pmatrix}}\,{\big |}\,2a-c-d=0{\text{ and }}a+3b=0\}$
* The subset $\{a+bx+cx^{3}\,{\big |}\,a-2b+c=0\}$ of ${\mathcal {P}}_{3}$
* The subset of ${\mathcal {P}}_{2}$ of quadratic polynomials $p$ such that $p(7)=0$

### Answer

* $\{{\begin{pmatrix}c&b&c\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}=\{b{\begin{pmatrix}0&1&0\end{pmatrix}}+c{\begin{pmatrix}1&0&1\end{pmatrix}}\,{\big |}\,b,c\in \mathbb {R} \}$ The obvious choice for the set that spans is $\{{\begin{pmatrix}0&1&0\end{pmatrix}},{\begin{pmatrix}1&0&1\end{pmatrix}}\}$.
* $\{{\begin{pmatrix}-d&b\\c&d\end{pmatrix}}\,{\big |}\,b,c,d\in \mathbb {R} \}=\{b{\begin{pmatrix}0&1\\0&0\end{pmatrix}}+c{\begin{pmatrix}0&0\\1&0\end{pmatrix}}+d{\begin{pmatrix}-1&0\\0&1\end{pmatrix}}\,{\big |}\,b,c,d\in \mathbb {R} \}$ One set that spans this space consists of those three matrices.
* The system gives $b=-(c+d)/6$ and $a=(c+d)/2$. So one description is this. That shows that a set spanning this subspace consists of those two matrices.
    ${\begin{array}{*{4}{rc}r}a&+&3b&&&&&=&0\\2a&&&&-c&-&d&=&0\end{array}}$
    $\{c{\begin{pmatrix}1/2&-1/6\\1&0\end{pmatrix}}+d{\begin{pmatrix}1/2&-1/6\\0&1\end{pmatrix}}\,{\big |}\,c,d\in \mathbb {R} \}$
* The $a=2b-c$ gives $\{(2b-c)+bx+cx^{3}\,{\big |}\,b,c\in \mathbb {R} \}=\{b(2+x)+c(-1+x^{3})\,{\big |}\,b,c\in \mathbb {R} \}$. So the subspace is the span of the set $\{2+x,-1+x^{3}\}$.
* The set $\{a+bx+cx^{2}\,{\big |}\,a+7b+49c=0\}$ parametrized as $\{b(-7+x)+c(-49+x^{2})\,{\big |}\,b,c\in \mathbb {R} \}$ has the spanning set $\{-7+x,-49+x^{2}\}$.
    This exercise is recommended for all readers.

### Problem 7

Find a set to span the given subspace of the given space. (Hint. Parametrize each.)
* the $xz$-plane in $\mathbb {R} ^{3}$
* $\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,3x+2y+z=0\}$ in $\mathbb {R} ^{3}$
* $\{{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\,{\big |}\,2x+y+w=0{\text{ and }}y+2z=0\}$ in $\mathbb {R} ^{4}$
* $\{a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\,{\big |}\,a_{0}+a_{1}=0{\text{ and }}a_{2}-a_{3}=0\}$ in ${\mathcal {P}}_{3}$
* The set ${\mathcal {P}}_{4}$ in the space ${\mathcal {P}}_{4}$
* ${\mathcal {M}}_{2\!\times \!2}$ in ${\mathcal {M}}_{2\!\times \!2}$

### Answer

Each answer given is only one out of many possible.
* We can parametrize in this way giving this for a spanning set.
    $\{{\begin{pmatrix}x\\0\\z\end{pmatrix}}\,{\big |}\,x,z\in \mathbb {R} \}=\{x{\begin{pmatrix}1\\0\\0\end{pmatrix}}+z{\begin{pmatrix}0\\0\\1\end{pmatrix}}\,{\big |}\,x,z\in \mathbb {R} \}$
    $\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\}$
* As a way to solve it, is to express $x$ as $-(2/3)y-(1/3)z$ to get this parametrization. and get this as a spanning set.
    $\{{\begin{pmatrix}-(2/3)y-(1/3)z\\y\\z\end{pmatrix}}\,{\big |}\,y,z\in \mathbb {R} \}=\{y{\begin{pmatrix}-2/3\\1\\0\end{pmatrix}}+z{\begin{pmatrix}-1/3\\0\\1\end{pmatrix}}\,{\big |}\,y,z\in \mathbb {R} \}$
    $\{{\begin{pmatrix}-2/3\\1\\0\end{pmatrix}},{\begin{pmatrix}-1/3\\0\\1\end{pmatrix}}\}$
* Here as a way to go, is to notice that $x$ and $z$ depend on $y$ and $w$, but not each other, suggesting to choose $y$ and $w$ as free variables. Then re-expressing, $x=-(1/2)y-(1/2)w$ and $z=-(1/2)y$ to obtain this parametrization. and for a possible spanning set we have.
    $\{{\begin{pmatrix}-(1/2)y-(1/2)w\\y\\-(1/2)y\\w\end{pmatrix}}\,{\big |}\,y,w\in \mathbb {R} \}=\{y{\begin{pmatrix}-1/2\\1\\-1/2\\0\end{pmatrix}}+w{\begin{pmatrix}-1/2\\0\\0\\1\end{pmatrix}}\,{\big |}\,y,w\in \mathbb {R} \}$
    $\{{\begin{pmatrix}-1/2\\1\\-1/2\\0\end{pmatrix}},{\begin{pmatrix}-1/2\\0\\0\\1\end{pmatrix}}\}$
* Again, as a way to do it, re-express $a_{0}=-a_{1}$ and $a_{2}=a_{3}$ to get parametrization. which gives this for a spanning set.
    $\{-a_{1}+a_{1}x+a_{3}x^{2}+a_{3}x^{3}\,{\big |}\,a_{1},a_{3}\in \mathbb {R} \}=\{a_{1}(-1+x)+a_{3}(x^{2}+x^{3})\,{\big |}\,a_{1},a_{3}\in \mathbb {R} \}\$
    $\{-1+x,\,x^{2}+x^{3}\}$
* As there are no restrictions, we can get a spaning set right away, skipping parametrization.
    $\{1,\,x,\,x^{2},\,x^{3},\,x^{4}\}$
* For same reason as the previous subproblem, we can skip parametrization and go to a spaning set immediatedly.
    $\{{\begin{pmatrix}1&0\\0&0\end{pmatrix}},{\begin{pmatrix}0&1\\0&0\end{pmatrix}},{\begin{pmatrix}0&0\\1&0\end{pmatrix}},{\begin{pmatrix}0&0\\0&1\end{pmatrix}}\}$

### Problem 8

Is $\mathbb {R} ^{2}$ a subspace of $\mathbb {R} ^{3}$?

### Answer

Technically, no. Subspaces of $\mathbb {R} ^{3}$ are sets of three-tall vectors, while $\mathbb {R} ^{2}$ is a set of two-tall vectors. Clearly though, $\mathbb {R} ^{2}$ is "just like" this subspace of $\mathbb {R} ^{3}$.
    $\{{\begin{pmatrix}x\\y\\0\end{pmatrix}}\,{\big |}\,x,y\in \mathbb {R} \}$
    This exercise is recommended for all readers.

### Problem 9

Decide if each is a subspace of the vector space of real-valued functions of one real variable.
* The even functions $\{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,f(-x)=f(x){\text{ for all }}x\}$. For example, two members of this set are $f_{1}(x)=x^{2}$ and $f_{2}(x)=\cos(x)$.
* The odd functions $\{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,f(-x)=-f(x){\text{ for all }}x\}$. Two members are $f_{3}(x)=x^{3}$ and $f_{4}(x)=\sin(x)$.

### Answer

Of course, the addition and scalar multiplication operations are the ones inherited from the enclosing space.
* This is a subspace. It is not empty as it contains at least the two example functions given. It is closed because if $f_{1},f_{2}$ are even and $c_{1},c_{2}$ are scalars then we have this.
    $(c_{1}f_{1}+c_{2}f_{2})\,(-x)=c_{1}\,f_{1}(-x)+c_{2}\,f_{2}(-x)=c_{1}\,f_{1}(x)+c_{2}\,f_{2}(x)=(c_{1}f_{1}+c_{2}f_{2})\,(x)$
* This is also a subspace; the check is similar to the prior one.

### Problem 10

Example 2.16 says that for any vector ${\vec {v}}$ that is an element of a vector space $V$, the set $\{r\cdot {\vec {v}}\,{\big |}\,r\in \mathbb {R} \}$ is a subspace of $V$. (This is of course, simply the span of the singleton set $\{{\vec {v}}\}$.) Must any such subspace be a proper subspace, or can it be improper?

### Answer

It can be improper. If ${\vec {v}}={\vec {0}}$ then this is a trivial subspace. At the opposite extreme, if the vector space is $\mathbb {R} ^{1}$ and ${\vec {v}}\neq {\vec {0}}\,$ then the subspace is all of $\mathbb {R} ^{1}$.

### Problem 11

An example following the definition of a vector space shows that the solution set of a homogeneous linear system is a vector space. In the terminology of this subsection, it is a subspace of $\mathbb {R} ^{n}$ where the system has $n$ variables. What about a non-homogeneous linear system; do its solutions form a subspace (under the inherited operations)?

### Answer

No, such a set is not closed. For one thing, it does not contain the zero vector.

### Problem 12

Example 2.19 shows that $\mathbb {R} ^{3}$ has infinitely many subspaces. Does every nontrivial space have infinitely many subspaces?

### Answer

No. The only subspaces of $\mathbb {R} ^{1}$ are the space itself and its trivial subspace. Any subspace $S$ of $\mathbb {R}$ that contains a nonzero member ${\vec {v}}$ must contain the set of all of its scalar multiples $\{r\cdot {\vec {v}}\,{\big |}\,r\in \mathbb {R} \}$. But this set is all of $\mathbb {R}$.

### Problem 13

Finish the proof of Lemma 2.9.

### Answer

Item (1) is checked in the text.
Item (2) has five conditions. First, for closure, if $c\in \mathbb {R}$ and ${\vec {s}}\in S$ then $c\cdot {\vec {s}}\in S$ as $c\cdot {\vec {s}}=c\cdot {\vec {s}}+0\cdot {\vec {0}}$. Second, because the operations in $S$ are inherited from $V$, for $c,d\in \mathbb {R}$ and ${\vec {s}}\in S$, the scalar product $(c+d)\cdot {\vec {s}}\,$ in $S$ equals the product $(c+d)\cdot {\vec {s}}\,$ in $V$, and that equals $c\cdot {\vec {s}}+d\cdot {\vec {s}}\,$ in $V$, which equals $c\cdot {\vec {s}}+d\cdot {\vec {s}}\,$ in $S$.
The check for the third, fourth, and fifth conditions are similar to the second conditions's check just given.

### Problem 14

Show that each vector space has only one trivial subspace.

### Answer

An exercise in the prior subsection shows that every vector space has only one zero vector (that is, there is only one vector that is the additive identity element of the space). But a trivial space has only one element and that element must be this (unique) zero vector.
    This exercise is recommended for all readers.

### Problem 15

Show that for any subset $S$ of a vector space, the span of the span equals the span $[[S]]=[S]$. (Hint. Members of $[S]$ are linear combinations of members of $S$. Members of $[[S]]$ are linear combinations of linear combinations of members of $S$.)

### Answer

As the hint suggests, the basic reason is the Linear Combination Lemma from the first chapter. For the full proof, we will show mutual containment between the two sets.
The first containment $[[S]]\supseteq [S]$ is an instance of the more general, and obvious, fact that for any subset $T$ of a vector space, $[T]\supseteq T$.
For the other containment, that $[[S]]\subseteq [S]$, take $m$ vectors from $[S]$, namely $c_{1,1}{\vec {s}}_{1,1}+\cdots +c_{1,n_{1}}{\vec {s}}_{1,n_{1}}$, ..., $c_{1,m}{\vec {s}}_{1,m}+\cdots +c_{1,n_{m}}{\vec {s}}_{1,n_{m}}$, and note that any linear combination of those
    $r_{1}(c_{1,1}{\vec {s}}_{1,1}+\cdots +c_{1,n_{1}}{\vec {s}}_{1,n_{1}})+\cdots +r_{m}(c_{1,m}{\vec {s}}_{1,m}+\cdots +c_{1,n_{m}}{\vec {s}}_{1,n_{m}})$
is a linear combination of elements of $S$
    $=(r_{1}c_{1,1}){\vec {s}}_{1,1}+\cdots +(r_{1}c_{1,n_{1}}){\vec {s}}_{1,n_{1}}+\cdots +(r_{m}c_{1,m}){\vec {s}}_{1,m}+\cdots +(r_{m}c_{1,n_{m}}){\vec {s}}_{1,n_{m}}$
and so is in $[S]$. That is, simply recall that a linear combination of linear combinations (of members of $S$) is a linear combination (again of members of $S$).

### Problem 16

All of the subspaces that we've seen use zero in their description in some way. For example, the subspace in Example 2.3 consists of all the vectors from $\mathbb {R} ^{2}$ with a second component of zero. In contrast, the collection of vectors from $\mathbb {R} ^{2}$ with a second component of one does not form a subspace (it is not closed under scalar multiplication). Another example is Example 2.2, where the condition on the vectors is that the three components add to zero. If the condition were that the three components add to one then it would not be a subspace (again, it would fail to be closed). This exercise shows that a reliance on zero is not strictly necessary. Consider the set
    $\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x+y+z=1\}$
under these operations.
    ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}-1\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}\qquad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rx-r+1\\ry\\rz\end{pmatrix}}$
* Show that it is not a subspace of $\mathbb {R} ^{3}$. (Hint. See Example 2.5).
* Show that it is a vector space. Note that by the prior item, Lemma 2.9 can not apply.
* Show that any subspace of $\mathbb {R} ^{3}$ must pass through the origin, and so any subspace of $\mathbb {R} ^{3}$ must involve zero in its description. Does the converse hold? Does any subset of $\mathbb {R} ^{3}$ that contains the origin become a subspace when given the inherited operations?

### Answer

* It is not a subspace because these are not the inherited operations. For one thing, in this space, while this does not, of course, hold in $\mathbb {R} ^{3}$.
    $0\cdot {\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$
* We can combine the argument showing closure under addition with the argument showing closure under scalar multiplication into one single argument showing closure under linear combinations of two vectors. If $r_{1},r_{2},x_{1},x_{2},y_{1},y_{2},z_{1},z_{2}$ are in $\mathbb {R}$ then (note that the definition of addition in this space is that the first components combine as $(r_{1}x_{1}-r_{1}+1)+(r_{2}x_{2}-r_{2}+1)-1$, so the first component of the last vector does not say "$+2$"). Adding the three components of the last vector gives $r_{1}(x_{1}-1+y_{1}+z_{1})+r_{2}(x_{2}-1+y_{2}+z_{2})+1=r_{1}\cdot 0+r_{2}\cdot 0+1=1$. Most of the other checks of the conditions are easy (although the oddness of the operations keeps them from being routine). Commutativity of addition goes like this. Associativity of addition has while and they are equal. The identity element with respect to this addition operation works this way and the additive inverse is similar. The conditions on scalar multiplication are also easy. For the first condition, while and the two are equal. The second condition compares with and they are equal. For the third condition, while and the two are equal. For scalar multiplication by $1$ we have this. Thus all the conditions on a vector space are met by these two operations. Remark. A way to understand this vector space is to think of it as the plane in $\mathbb {R} ^{3}$ displaced away from the origin by $1$ along the $x$-axis. Then addition becomes: to add two members of this space, (such that $x_{1}+y_{1}+z_{1}=1$ and $x_{2}+y_{2}+z_{2}=1$) move them back by $1$ to place them in $P$ and add as usual, and then move the result back out by $1$ along the $x$-axis. Scalar multiplication is similar.
    $r_{1}{\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+r_{2}{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}r_{1}x_{1}-r_{1}+1\\r_{1}y_{1}\\r_{1}z_{1}\end{pmatrix}}+{\begin{pmatrix}r_{2}x_{2}-r_{2}+1\\r_{2}y_{2}\\r_{2}z_{2}\end{pmatrix}}={\begin{pmatrix}r_{1}x_{1}-r_{1}+r_{2}x_{2}-r_{2}+1\\r_{1}y_{1}+r_{2}y_{2}\\r_{1}z_{1}+r_{2}z_{2}\end{pmatrix}}$
    ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}-1\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}={\begin{pmatrix}x_{2}+x_{1}-1\\y_{2}+y_{1}\\z_{2}+z_{1}\end{pmatrix}}={\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}+{\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}$
    $({\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}})+{\begin{pmatrix}x_{3}\\y_{3}\\z_{3}\end{pmatrix}}={\begin{pmatrix}(x_{1}+x_{2}-1)+x_{3}-1\\(y_{1}+y_{2})+y_{3}\\(z_{1}+z_{2})+z_{3}\end{pmatrix}}$
    ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+({\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}+{\begin{pmatrix}x_{3}\\y_{3}\\z_{3}\end{pmatrix}})={\begin{pmatrix}x_{1}+(x_{2}+x_{3}-1)-1\\y_{1}+(y_{2}+y_{3})\\z_{1}+(z_{2}+z_{3})\end{pmatrix}}$
    ${\begin{pmatrix}x\\y\\z\end{pmatrix}}+{\begin{pmatrix}1\\0\\0\end{pmatrix}}={\begin{pmatrix}x+1-1\\y+0\\z+0\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    ${\begin{pmatrix}x\\y\\z\end{pmatrix}}+{\begin{pmatrix}-x+2\\-y\\-z\end{pmatrix}}={\begin{pmatrix}x+(-x+2)-1\\y-y\\z-z\end{pmatrix}}={\begin{pmatrix}1\\0\\0\end{pmatrix}}$
    $(r+s){\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}(r+s)x-(r+s)+1\\(r+s)y\\(r+s)z\end{pmatrix}}$
    $r{\begin{pmatrix}x\\y\\z\end{pmatrix}}+s{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rx-r+1\\ry\\rz\end{pmatrix}}+{\begin{pmatrix}sx-s+1\\sy\\sz\end{pmatrix}}={\begin{pmatrix}(rx-r+1)+(sx-s+1)-1\\ry+sy\\rz+sz\end{pmatrix}}$
    $r\cdot ({\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}})=r\cdot {\begin{pmatrix}x_{1}+x_{2}-1\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}={\begin{pmatrix}r(x_{1}+x_{2}-1)-r+1\\r(y_{1}+y_{2})\\r(z_{1}+z_{2})\end{pmatrix}}$
    $r{\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+r{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}rx_{1}-r+1\\ry_{1}\\rz_{1}\end{pmatrix}}+{\begin{pmatrix}rx_{2}-r+1\\ry_{2}\\rz_{2}\end{pmatrix}}={\begin{pmatrix}(rx_{1}-r+1)+(rx_{2}-r+1)-1\\ry_{1}+ry_{2}\\rz_{1}+rz_{2}\end{pmatrix}}$
    $(rs){\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rsx-rs+1\\rsy\\rsz\end{pmatrix}}$
    $r(s{\begin{pmatrix}x\\y\\z\end{pmatrix}})=r({\begin{pmatrix}sx-s+1\\sy\\sz\end{pmatrix}})={\begin{pmatrix}r(sx-s+1)-r+1\\rsy\\rsz\end{pmatrix}}$
    $1\cdot {\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}1x-1+1\\1y\\1z\end{pmatrix}}={\begin{pmatrix}x\\y\\z\end{pmatrix}}$
    $P=\{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\,{\big |}\,x+y+z=0\}$
    ${\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}},\;{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}$
    ${\begin{pmatrix}x_{1}-1\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}-1\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}-2\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}\qquad {\text{(in }}P{\text}}$
    ${\begin{pmatrix}x_{1}+x_{2}-1\\y_{1}+y_{2}\\z_{1}+z_{2}\end{pmatrix}}.$
* For the subspace to be closed under the inherited scalar multiplication, where ${\vec {v}}$ is a member of that subspace, must also be a member. The converse does not hold. Here is a subset of $\mathbb {R} ^{3}$ that contains the origin (this subset has only two elements) but is not a subspace.
    $0\cdot {\vec {v}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}$
    $\{{\begin{pmatrix}0\\0\\0\end{pmatrix}},{\begin{pmatrix}1\\0\\0\end{pmatrix}}\}$

### Problem 17

We can give a justification for the convention that the sum of zero-many vectors equals the zero vector. Consider this sum of three vectors ${\vec {v}}_{1}+{\vec {v}}_{2}+{\vec {v}}_{3}$.
* What is the difference between this sum of three vectors and the sum of the first two of these three?
* What is the difference between the prior sum and the sum of just the first one vector?
* What should be the difference between the prior sum of one vector and the sum of no vectors?
* So what should be the definition of the sum of no vectors?

### Answer

* $({\vec {v}}_{1}+{\vec {v}}_{2}+{\vec {v}}_{3})-({\vec {v}}_{1}+{\vec {v}}_{2})={\vec {v}}_{3}$
* $({\vec {v}}_{1}+{\vec {v}}_{2})-({\vec {v}}_{1})={\vec {v}}_{2}$
* Surely, ${\vec {v}}_{1}$.
* Taking the one-long sum and subtracting gives (${\vec {v}}_{1})-{\vec {v}}_{1}={\vec {0}}$.

### Problem 18

Is a space determined by its subspaces? That is, if two vector spaces have the same subspaces, must the two be equal?

### Answer

Yes; any space is a subspace of itself, so each space contains the other.

### Problem 19

* Give a set that is closed under scalar multiplication but not addition.
* Give a set closed under addition but not scalar multiplication.
* Give a set closed under neither.

### Answer

* The union of the $x$-axis and the $y$-axis in $\mathbb {R} ^{2}$ is one.
* The set of integers, as a subset of $\mathbb {R} ^{1}$, is one.
* The subset $\{{\vec {v}}\}$ of $\mathbb {R} ^{2}$ is one, where ${\vec {v}}$ is any nonzero vector.

### Problem 20

Show that the span of a set of vectors does not depend on the order in which the vectors are listed in that set.

### Answer

Because vector space addition is commutative, a reordering of summands leaves a linear combination unchanged.

### Problem 21

Which trivial subspace is the span of the empty set? Is it
    $\{{\begin{pmatrix}0\\0\\0\end{pmatrix}}\}\subseteq \mathbb {R} ^{3},\quad {\text{or}}\quad \{0+0x\}\subseteq {\mathcal {P}}_{1},$
or some other subspace?

### Answer

We always consider that span in the context of an enclosing space.

### Problem 22

Show that if a vector is in the span of a set then adding that vector to the set won't make the span any bigger. Is that also "only if"?

### Answer

It is both "if" and "only if".
For "if", let $S$ be a subset of a vector space $V$ and assume ${\vec {v}}\in S$ satisfies ${\vec {v}}=c_{1}{\vec {s}}_{1}+\dots +c_{n}{\vec {s}}_{n}$ where $c_{1},\ldots ,c_{n}$ are scalars and ${\vec {s}}_{1},\ldots ,{\vec {s}}_{n}\in S$. We must show that $[S\cup \{{\vec {v}}\}]=[S]$.
Containment one way, $[S]\subseteq [S\cup \{{\vec {v}}\}]$ is obvious. For the other direction, $[S\cup \{{\vec {v}}\}]\subseteq [S]$, note that if a vector is in the set on the left then it has the form $d_{0}{\vec {v}}+d_{1}{\vec {t}}_{1}+\dots +d_{m}{\vec {t}}_{m}$ where the $d$'s are scalars and the ${\vec {t}}\,$'s are in $S$. Rewrite that as $d_{0}(c_{1}{\vec {s}}_{1}+\dots +c_{n}{\vec {s}}_{n})+d_{1}{\vec {t}}_{1}+\cdots +d_{m}{\vec {t}}_{m}$ and note that the result is a member of the span of $S$.
The "only if" is clearly true— adding ${\vec {v}}$ enlarges the span to include at least ${\vec {v}}$.
    This exercise is recommended for all readers.

### Problem 23

Subspaces are subsets and so we naturally consider how "is a subspace of" interacts with the usual set operations.
* If $A,B$ are subspaces of a vector space, must $A\cap B$ be a subspace? Always? Sometimes? Never?
* Must $A\cup B$ be a subspace?
* If $A$ is a subspace, must its complement be a subspace?
(Hint. Try some test subspaces from Example 2.19.)

### Answer

* Always. Assume that $A,B$ are subspaces of $V$. Note that their intersection is not empty as both contain the zero vector. If ${\vec {w}},{\vec {v}}\in A\cap B$ and $r,s$ are scalars then $r{\vec {v}}+s{\vec {w}}\in A$ because each vector is in $A$ and so a linear combination is in $A$, and $r{\vec {v}}+s{\vec {w}}\in B$ for the same reason. Thus the intersection is closed. Now Lemma 2.19 applies.
* Sometimes (more precisely, only if $A\subseteq B$ or $B\subseteq A$). To see the answer is not "always", take $V$ to be $\mathbb {R} ^{2}$, take $A$ to be the $x$-axis, and $B$ to be the $y$-axis. Note that as the sum is in neither $A$ nor $B$. The answer is not "never" because if $A\subseteq B$ or $B\subseteq A$ then clearly $A\cup B$ is a subspace. To show that $A\cup B$ is a subspace only if one subspace contains the other, we assume that $A\not \subseteq B$ and $B\not \subseteq A$ and prove that the union is not a subspace. The assumption that $A$ is not a subset of $B$ means that there is an ${\vec {a}}\in A$ with ${\vec {a}}\not \in B$. The other assumption gives a ${\vec {b}}\in B$ with ${\vec {b}}\not \in A$. Consider ${\vec {a}}+{\vec {b}}$. Note that sum is not an element of $A$ or else $({\vec {a}}+{\vec {b}})-{\vec {a}}$ would be in $A$, which it is not. Similarly the sum is not an element of $B$. Hence the sum is not an element of $A\cup B$, and so the union is not a subspace.
    ${\begin{pmatrix}1\\0\end{pmatrix}}\in A{\text{ and }}{\begin{pmatrix}0\\1\end{pmatrix}}\in B\quad {\text{but}}\quad {\begin{pmatrix}1\\0\end{pmatrix}}+{\begin{pmatrix}0\\1\end{pmatrix}}\not \in A\cup B$
* Never. As $A$ is a subspace, it contains the zero vector, and therefore the set that is $A$'s complement does not. Without the zero vector, the complement cannot be a vector space.
    This exercise is recommended for all readers.

### Problem 24

Does the span of a set depend on the enclosing space? That is, if $W$ is a subspace of $V$ and $S$ is a subset of $W$ (and so also a subset of $V$), might the span of $S$ in $W$ differ from the span of $S$ in $V$?

### Answer

The span of a set does not depend on the enclosing space. A linear combination of vectors from $S$ gives the same sum whether we regard the operations as those of $W$ or as those of $V$, because the operations of $W$ are inherited from $V$.

### Problem 25

Is the relation "is a subspace of" transitive? That is, if $V$ is a subspace of $W$ and $W$ is a subspace of $X$, must $V$ be a subspace of $X$?

### Answer

It is; apply Lemma 2.19. (You must consider the following. Suppose $B$ is a subspace of a vector space $V$ and suppose $A\subseteq B\subseteq V$ is a subspace. From which space does $A$ inherit its operations? The answer is that it doesn't matter— $A$ will inherit the same operations in either case.)
    This exercise is recommended for all readers.

### Problem 26

Because "span of" is an operation on sets we naturally consider how it interacts with the usual set operations.
* If $S\subseteq T$ are subsets of a vector space, is $[S]\subseteq [T]$? Always? Sometimes? Never?
* If $S,T$ are subsets of a vector space, is $[S\cup T]=[S]\cup [T]$?
* If $S,T$ are subsets of a vector space, is $[S\cap T]=[S]\cap [T]$?
* Is the span of the complement equal to the complement of the span?

### Answer

* Always; if $S\subseteq T$ then a linear combination of elements of $S$ is also a linear combination of elements of $T$.
* Sometimes (more precisely, if and only if $S\subseteq T$ or $T\subseteq S$). The answer is not "always" as is shown by this example from $\mathbb {R} ^{3}$ because of this. The answer is not "never" because if either set contains the other then equality is clear. We can characterize equality as happening only when either set contains the other by assuming $S\not \subseteq T$ (implying the existence of a vector ${\vec {s}}\in S$ with ${\vec {s}}\not \in T$) and $T\not \subseteq S$ (giving a ${\vec {t}}\in T$ with ${\vec {t}}\not \in S$), noting ${\vec {s}}+{\vec {t}}\in [S\cup T]$, and showing that ${\vec {s}}+{\vec {t}}\not \in [S]\cup [T]$.
    $S=\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\},\quad T=\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\}$
    ${\begin{pmatrix}1\\1\\1\end{pmatrix}}\in [S\cup T]\qquad {\begin{pmatrix}1\\1\\1\end{pmatrix}}\not \in [S]\cup [T]$
* Sometimes. Clearly $[S\cap T]\subseteq [S]\cap [T]$ because any linear combination of vectors from $S\cap T$ is a combination of vectors from $S$ and also a combination of vectors from $T$. Containment the other way does not always hold. For instance, in $\mathbb {R} ^{2}$, take so that $[S]\cap [T]$ is the $x$-axis but $[S\cap T]$ is the trivial subspace. Characterizing exactly when equality holds is tough. Clearly equality holds if either set contains the other, but that is not "only if" by this example in $\mathbb {R} ^{3}$.
    $S=\{{\begin{pmatrix}1\\0\end{pmatrix}},{\begin{pmatrix}0\\1\end{pmatrix}}\},\quad T=\{{\begin{pmatrix}2\\0\end{pmatrix}}\}$
    $S=\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\},\quad T=\{{\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\0\\1\end{pmatrix}}\}$
* Never, as the span of the complement is a subspace, while the complement of the span is not (it does not contain the zero vector).

### Problem 27

Reprove Lemma 2.15 without doing the empty set separately.

### Answer

Call the subset $S$. By Lemma 2.9, we need to check that $[S]$ is closed under linear combinations. If $c_{1}{\vec {s}}_{1}+\dots +c_{n}{\vec {s}}_{n},c_{n+1}{\vec {s}}_{n+1}+\dots +c_{m}{\vec {s}}_{m}\in [S]$ then for any $p,r\in \mathbb {R}$ we have
    $p\cdot (c_{1}{\vec {s}}_{1}+\cdots +c_{n}{\vec {s}}_{n})+r\cdot (c_{n+1}{\vec {s}}_{n+1}+\cdots +c_{m}{\vec {s}}_{m})=pc_{1}{\vec {s}}_{1}+\cdots +pc_{n}{\vec {s}}_{n}+rc_{n+1}{\vec {s}}_{n+1}+\cdots +rc_{m}{\vec {s}}_{m}$
which is an element of $[S]$. (Remark. If the set $S$ is empty, then that "if ... then ..." statement is vacuously true.)

### Problem 28

Find a structure that is closed under linear combinations, and yet is not a vector space. (Remark. This is a bit of a trick question.)

### Answer

For this to happen, one of the conditions giving the sensibleness of the addition and scalar multiplication operations must be violated. Consider $\mathbb {R} ^{2}$ with these operations.
    ${\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}\qquad r{\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}0\\0\end{pmatrix}}$
The set $\mathbb {R} ^{2}$ is closed under these operations. But it is not a vector space.
    $1\cdot {\begin{pmatrix}1\\1\end{pmatrix}}\neq {\begin{pmatrix}1\\1\end{pmatrix}}$