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

## Solutions

Problem 1

Name the zero vector for each of these vector spaces.

1. The space of degree three polynomials under the natural operations
2. The space of ${\displaystyle 2\!\times \!4}$  matrices
3. The space ${\displaystyle \{f:[0,1]\to \mathbb {R} \,{\big |}\,f{\text{ is continuous}}\}}$
4. The space of real-valued functions of one natural number variable
1. ${\displaystyle 0+0x+0x^{2}+0x^{3}}$
2. ${\displaystyle {\begin{pmatrix}0&0&0&0\\0&0&0&0\end{pmatrix}}}$
3. The constant function ${\displaystyle f(x)=0}$
4. The constant function ${\displaystyle f(n)=0}$
This exercise is recommended for all readers.
Problem 2

Find the additive inverse, in the vector space, of the vector.

1. In ${\displaystyle {\mathcal {P}}_{3}}$ , the vector ${\displaystyle -3-2x+x^{2}}$ .
2. In the space ${\displaystyle 2\!\times \!2}$ ,
${\displaystyle {\begin{pmatrix}1&-1\\0&3\end{pmatrix}}.}$
3. In ${\displaystyle \{ae^{x}+be^{-x}\,{\big |}\,a,b\in \mathbb {R} \}}$ , the space of functions of the real variable ${\displaystyle x}$  under the natural operations, the vector ${\displaystyle 3e^{x}-2e^{-x}}$ .
1. ${\displaystyle 3+2x-x^{2}}$
2. ${\displaystyle {\begin{pmatrix}-1&+1\\0&-3\end{pmatrix}}}$
3. ${\displaystyle -3e^{x}+2e^{-x}}$
This exercise is recommended for all readers.
Problem 3

Show that each of these is a vector space.

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

Most of the conditions are easy to check; use Example 1.3 as a guide. Here are some comments.

1. This is just like Example 1.3; the zero element is ${\displaystyle 0+0x}$ .
2. The zero element of this space is the ${\displaystyle 2\!\times \!2}$  matrix of zeroes.
3. The zero element is the vector of zeroes.
4. Closure of addition involves noting that the sum
${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\\w_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\\w_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}+x_{2}\\y_{1}+y_{2}\\z_{1}+z_{2}\\w_{1}+w_{2}\end{pmatrix}}}$
is in ${\displaystyle L}$  because ${\displaystyle (x_{1}+x_{2})+(y_{1}+y_{2})-(z_{1}+z_{2})+(w_{1}+w_{2})=(x_{1}+y_{1}-z_{1}+w_{1})+(x_{2}+y_{2}-z_{2}+w_{2})=0+0}$ . Closure of scalar multiplication is similar. Note that the zero element, the vector of zeroes, is in ${\displaystyle L}$ .
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.)

1. Under the operations inherited from ${\displaystyle \mathbb {R} ^{3}}$ , this set
${\displaystyle \{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\,{\big |}\,x+y+z=1\}}$
2. Under the operations inherited from ${\displaystyle \mathbb {R} ^{3}}$ , this set
${\displaystyle \{{\begin{pmatrix}x\\y\\z\end{pmatrix}}\in \mathbb {R} ^{3}\,{\big |}\,x^{2}+y^{2}+z^{2}=1\}}$
3. Under the usual matrix operations,
${\displaystyle \{{\begin{pmatrix}a&1\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}}$
4. Under the usual polynomial operations,
${\displaystyle \{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0},a_{1},a_{2}\in \mathbb {R} ^{+}\}}$
where ${\displaystyle \mathbb {R} ^{+}}$  is the set of reals greater than zero
5. Under the inherited operations,
${\displaystyle \{{\begin{pmatrix}x\\y\end{pmatrix}}\in \mathbb {R} ^{2}\,{\big |}\,x+3y=4{\text{ and }}2x-y=3{\text{ and }}6x+4y=10\}}$

In each item the set is called ${\displaystyle Q}$ . For some items, there are other correct ways to show that ${\displaystyle Q}$  is not a vector space.

1. It is not closed under addition; it fails to meet condition 1.
${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}\not \in Q}$
2. It is not closed under addition.
${\displaystyle {\begin{pmatrix}1\\0\\0\end{pmatrix}},{\begin{pmatrix}0\\1\\0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1\\1\\0\end{pmatrix}}\not \in Q}$
3. It is not closed under addition.
${\displaystyle {\begin{pmatrix}0&1\\0&0\end{pmatrix}},\,{\begin{pmatrix}1&1\\0&0\end{pmatrix}}\in Q\qquad {\begin{pmatrix}1&2\\0&0\end{pmatrix}}\not \in Q}$
4. It is not closed under scalar multiplication.
${\displaystyle 1+1x+1x^{2}\in Q\qquad -1\cdot (1+1x+1x^{2})\not \in Q}$
5. It is empty, violating condition 4.
Problem 5

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

The usual operations ${\displaystyle (v_{0}+v_{1}i)+(w_{0}+w_{1}i)=(v_{0}+w_{0})+(v_{1}+w_{1})i}$  and ${\displaystyle r(v_{0}+v_{1}i)=(rv_{0})+(rv_{1})i}$  suffice. The check is easy.

This exercise is recommended for all readers.
Problem 6

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

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

Problem 7

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

The natural operations are ${\displaystyle (v_{1}x+v_{2}y+v_{3}z)+(w_{1}x+w_{2}y+w_{3}z)=(v_{1}+w_{1})x+(v_{2}+w_{2})y+(v_{3}+w_{3})z}$  and ${\displaystyle r\cdot (v_{1}x+v_{2}y+v_{3}z)=(rv_{1})x+(rv_{2})y+(rv_{3})z}$ . 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.

${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\end{pmatrix}}={\begin{pmatrix}x_{1}-x_{2}\\y_{1}-y_{2}\end{pmatrix}}\qquad r\cdot {\begin{pmatrix}x\\y\end{pmatrix}}={\begin{pmatrix}rx\\ry\end{pmatrix}}}$

The "${\displaystyle +}$ " 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 ${\displaystyle \mathbb {R} ^{3}}$  is a vector space under these operations.

1. ${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\quad {\text{and}}\quad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}rx\\ry\\rz\end{pmatrix}}}$
2. ${\displaystyle {\begin{pmatrix}x_{1}\\y_{1}\\z_{1}\end{pmatrix}}+{\begin{pmatrix}x_{2}\\y_{2}\\z_{2}\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}\quad {\text{and}}\quad r{\begin{pmatrix}x\\y\\z\end{pmatrix}}={\begin{pmatrix}0\\0\\0\end{pmatrix}}}$
1. It is not a vector space.
${\displaystyle (1+1)\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\neq {\begin{pmatrix}1\\0\\0\end{pmatrix}}+{\begin{pmatrix}1\\0\\0\end{pmatrix}}}$
2. It is not a vector space.
${\displaystyle 1\cdot {\begin{pmatrix}1\\0\\0\end{pmatrix}}\neq {\begin{pmatrix}1\\0\\0\end{pmatrix}}}$
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.

1. The diagonal ${\displaystyle 2\!\times \!2}$  matrices
${\displaystyle \{{\begin{pmatrix}a&0\\0&b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}}$
2. This set of ${\displaystyle 2\!\times \!2}$  matrices
${\displaystyle \{{\begin{pmatrix}x&x+y\\x+y&y\end{pmatrix}}\,{\big |}\,x,y\in \mathbb {R} \}}$
3. This set
${\displaystyle \{{\begin{pmatrix}x\\y\\z\\w\end{pmatrix}}\in \mathbb {R} ^{4}\,{\big |}\,x+y+z+w=1\}}$
4. The set of functions ${\displaystyle \{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,df/dx+2f=0\}}$
5. The set of functions ${\displaystyle \{f:\mathbb {R} \to \mathbb {R} \,{\big |}\,df/dx+2f=1\}}$

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.

1. Yes.
2. Yes.
3. No, it is not closed under addition. The vector of all ${\displaystyle 1/4}$ 's, when added to itself, makes a nonmember.
4. Yes.
5. No, ${\displaystyle f(x)=e^{-2x}+(1/2)}$  is in the set but ${\displaystyle 2\cdot f}$  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 ${\displaystyle f}$  of one real variable such that ${\displaystyle f(7)=0}$ .

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

This exercise is recommended for all readers.
Problem 12

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

We check Definition 1.1.

First, closure under "${\displaystyle +}$ " 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 ${\displaystyle 1\in \mathbb {R} ^{+}}$  won't change the number. Fifth, any positive real has a reciprocal that is a positive real.

The sixth, closure under "${\displaystyle \cdot }$ ", holds because any power of a positive real is a positive real. The seventh condition is just the rule that ${\displaystyle v^{r+s}}$  equals the product of ${\displaystyle v^{r}}$  and ${\displaystyle v^{s}}$ . The eight condition says that ${\displaystyle (vw)^{r}=v^{r}w^{r}}$ . The ninth condition asserts that ${\displaystyle (v^{r})^{s}=v^{rs}}$ . The final condition says that ${\displaystyle v^{1}=v}$ .

Problem 13

Is ${\displaystyle \{(x,y)\,{\big |}\,x,y\in \mathbb {R} \}}$  a vector space under these operations?

1. ${\displaystyle (x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})}$  and ${\displaystyle r\cdot (x,y)=(rx,y)}$
2. ${\displaystyle (x_{1},y_{1})+(x_{2},y_{2})=(x_{1}+x_{2},y_{1}+y_{2})}$  and ${\displaystyle r\cdot (x,y)=(rx,0)}$
1. No: ${\displaystyle 1\cdot (0,1)+1\cdot (0,1)\neq (1+1)\cdot (0,1)}$ .
2. 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 ${\displaystyle 1\cdot (0,1)\neq (0,1)}$ .
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.

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

Problem 15

At this point "the same" is only an intuition, but nonetheless for each vector space identify the ${\displaystyle k}$  for which the space is "the same" as ${\displaystyle \mathbb {R} ^{k}}$ .

1. The ${\displaystyle 2\!\times \!3}$  matrices under the usual operations
2. The ${\displaystyle n\!\times \!m}$  matrices (under their usual operations)
3. This set of ${\displaystyle 2\!\times \!2}$  matrices
${\displaystyle \{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a,b,c\in \mathbb {R} \}}$
4. This set of ${\displaystyle 2\!\times \!2}$  matrices
${\displaystyle \{{\begin{pmatrix}a&0\\b&c\end{pmatrix}}\,{\big |}\,a+b+c=0\}}$
1. ${\displaystyle 6}$
2. ${\displaystyle nm}$
3. ${\displaystyle 3}$
4. To see that the answer is ${\displaystyle 2}$ , rewrite it as
${\displaystyle \{{\begin{pmatrix}a&0\\b&-a-b\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {R} \}}$
so that there are two parameters.
This exercise is recommended for all readers.
Problem 16

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

A vector space (over ${\displaystyle \mathbb {R} }$ ) consists of a set ${\displaystyle V}$  along with two operations "${\displaystyle {\mathbin {\vec {+}}}}$ " and "${\displaystyle {\mathbin {\vec {\cdot }}}}$ " subject to these conditions. Where ${\displaystyle {\vec {v}},{\vec {w}}\in V}$ ,

1. their vector sum ${\displaystyle {\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}}$  is an element of ${\displaystyle V}$ . If ${\displaystyle {\vec {u}},{\vec {v}},{\vec {w}}\in V}$  then
2. ${\displaystyle {\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}={\vec {w}}{\mathbin {\vec {+}}}{\vec {v}}}$  and
3. ${\displaystyle ({\vec {v}}{\mathbin {\vec {+}}}{\vec {w}}){\mathbin {\vec {+}}}{\vec {u}}={\vec {v}}{\mathbin {\vec {+}}}({\vec {w}}{\mathbin {\vec {+}}}{\vec {u}})}$ .
4. There is a zero vector ${\displaystyle {\vec {0}}\in V}$  such that ${\displaystyle {\vec {v}}{\mathbin {\vec {+}}}{\vec {0}}={\vec {v}}\,}$  for all ${\displaystyle {\vec {v}}\in V}$ .
5. Each ${\displaystyle {\vec {v}}\in V}$  has an additive inverse ${\displaystyle {\vec {w}}\in V}$  such that ${\displaystyle {\vec {w}}{\mathbin {\vec {+}}}{\vec {v}}={\vec {0}}}$ . If ${\displaystyle r,s}$  are scalars, that is, members of ${\displaystyle \mathbb {R} }$ ), and ${\displaystyle {\vec {v}},{\vec {w}}\in V}$  then
6. each scalar multiple ${\displaystyle r\cdot {\vec {v}}}$  is in ${\displaystyle V}$ . If ${\displaystyle r,s\in \mathbb {R} }$  and ${\displaystyle {\vec {v}},{\vec {w}}\in V}$  then
7. ${\displaystyle (r+s)\cdot {\vec {v}}=r\cdot {\vec {v}}{\mathbin {\vec {+}}}s\cdot {\vec {v}}}$ , and
8. ${\displaystyle r{\mathbin {\vec {\cdot }}}({\vec {v}}+{\vec {w}})=r{\mathbin {\vec {\cdot }}}{\vec {v}}+r{\mathbin {\vec {\cdot }}}{\vec {w}}}$ , and
9. ${\displaystyle (rs){\mathbin {\vec {\cdot }}}{\vec {v}}=r{\mathbin {\vec {\cdot }}}(s{\mathbin {\vec {\cdot }}}{\vec {v}})}$ , and
10. ${\displaystyle 1{\mathbin {\vec {\cdot }}}{\vec {v}}={\vec {v}}}$ .
This exercise is recommended for all readers.
Problem 17

Prove these.

1. Any vector is the additive inverse of the additive inverse of itself.
2. Vector addition left-cancels: if ${\displaystyle {\vec {v}},{\vec {s}},{\vec {t}}\in V}$  then ${\displaystyle {\vec {v}}+{\vec {s}}={\vec {v}}+{\vec {t}}\,}$  implies that ${\displaystyle {\vec {s}}={\vec {t}}}$ .
1. Let ${\displaystyle V}$  be a vector space, assume that ${\displaystyle {\vec {v}}\in V}$ , and assume that ${\displaystyle {\vec {w}}\in V}$  is the additive inverse of ${\displaystyle {\vec {v}}}$  so that ${\displaystyle {\vec {w}}+{\vec {v}}={\vec {0}}}$ . Because addition is commutative, ${\displaystyle {\vec {0}}={\vec {w}}+{\vec {v}}={\vec {v}}+{\vec {w}}}$ , so therefore ${\displaystyle {\vec {v}}}$  is also the additive inverse of ${\displaystyle {\vec {w}}}$ .
2. Let ${\displaystyle V}$  be a vector space and suppose ${\displaystyle {\vec {v}},{\vec {s}},{\vec {t}}\in V}$ . The additive inverse of ${\displaystyle {\vec {v}}}$  is ${\displaystyle -{\vec {v}}}$  so ${\displaystyle {\vec {v}}+{\vec {s}}={\vec {v}}+{\vec {t}}}$  gives that ${\displaystyle -{\vec {v}}+{\vec {v}}+{\vec {s}}=-{\vec {v}}+{\vec {v}}+{\vec {t}}}$ , which says that ${\displaystyle {\vec {0}}+{\vec {s}}={\vec {0}}+{\vec {t}}}$  and so ${\displaystyle {\vec {s}}={\vec {t}}}$ .
Problem 18

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

Addition is commutative, so in any vector space, for any vector ${\displaystyle {\vec {v}}}$  we have that ${\displaystyle {\vec {v}}={\vec {v}}+{\vec {0}}={\vec {0}}+{\vec {v}}}$ .

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.

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?

Each element of a vector space has one and only one additive inverse.

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

Problem 21
1. Prove that every point, line, or plane thru the origin in ${\displaystyle \mathbb {R} ^{3}}$  is a vector space under the inherited operations.
2. What if it doesn't contain the origin?
1. Every such set has the form ${\displaystyle \{r\cdot {\vec {v}}+s\cdot {\vec {w}}\,{\big |}\,r,s\in \mathbb {R} \}}$  where either or both of ${\displaystyle {\vec {v}},{\vec {w}}}$  may be ${\displaystyle {\vec {0}}}$ . With the inherited operations, closure of addition ${\displaystyle (r_{1}{\vec {v}}+s_{1}{\vec {w}})+(r_{2}{\vec {v}}+s_{2}{\vec {w}})=(r_{1}+r_{2}){\vec {v}}+(s_{1}+s_{2}){\vec {w}}}$  and scalar multiplication ${\displaystyle c(r{\vec {v}}+s{\vec {w}})=(cr){\vec {v}}+(cs){\vec {w}}}$  are easy. The other conditions are also routine.
2. 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 ${\displaystyle {\vec {v}}\in V}$  is not ${\displaystyle {\vec {0}}}$ .

1. Prove that ${\displaystyle r\cdot {\vec {v}}={\vec {0}}}$  if and only if ${\displaystyle r=0}$ .
2. Prove that ${\displaystyle r_{1}\cdot {\vec {v}}=r_{2}\cdot {\vec {v}}}$  if and only if ${\displaystyle r_{1}=r_{2}}$ .
3. Prove that any nontrivial vector space is infinite.
4. Use the fact that a nonempty solution set of a homogeneous linear system is a vector space to draw the conclusion.

Assume that ${\displaystyle {\vec {v}}\in V}$  is not ${\displaystyle {\vec {0}}}$ .

1. One direction of the if and only if is clear: if ${\displaystyle r=0}$  then ${\displaystyle r\cdot {\vec {v}}={\vec {0}}}$ . For the other way, let ${\displaystyle r}$  be a nonzero scalar. If ${\displaystyle r{\vec {v}}={\vec {0}}}$  then ${\displaystyle 1\cdot {\vec {v}}=(r/r)\cdot {\vec {v}}=(1/r)\cdot r{\vec {v}}=(1/r)\cdot {\vec {0}}}$  shows that ${\displaystyle {\vec {v}}={\vec {0}}}$ , contrary to the assumption.
2. Where ${\displaystyle r_{1},r_{2}}$  are scalars, ${\displaystyle r_{1}{\vec {v}}=r_{2}{\vec {v}}\,}$  holds if and only if ${\displaystyle (r_{1}-r_{2}){\vec {v}}={\vec {0}}}$ . By the prior item, then ${\displaystyle r_{1}-r_{2}=0}$ .
3. A nontrivial space has a vector ${\displaystyle {\vec {v}}\neq {\vec {0}}}$ . Consider the set ${\displaystyle \{k\cdot {\vec {v}}\,{\big |}\,k\in \mathbb {R} \}}$ . By the prior item this set is infinite.
4. 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?

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

Problem 24

A vector space over the complex numbers ${\displaystyle \mathbb {C} }$  has the same definition as a vector space over the reals except that scalars are drawn from ${\displaystyle \mathbb {C} }$  instead of from ${\displaystyle \mathbb {R} }$ . Show that each of these is a vector space over the complex numbers. (Recall how complex numbers add and multiply: ${\displaystyle (a_{0}+a_{1}i)+(b_{0}+b_{1}i)=(a_{0}+b_{0})+(a_{1}+b_{1})i}$  and ${\displaystyle (a_{0}+a_{1}i)(b_{0}+b_{1}i)=(a_{0}b_{0}-a_{1}b_{1})+(a_{0}b_{1}+a_{1}b_{0})i}$ .)

1. The set of degree two polynomials with complex coefficients
2. This set
${\displaystyle \{{\begin{pmatrix}0&a\\b&0\end{pmatrix}}\,{\big |}\,a,b\in \mathbb {C} {\text{ and }}a+b=0+0i\}}$

The check is routine. Note that "1" is ${\displaystyle 1+0i}$  and the zero elements are these.

1. ${\displaystyle (0+0i)+(0+0i)x+(0+0i)x^{2}}$
2. ${\displaystyle {\begin{pmatrix}0+0i&0+0i\\0+0i&0+0i\end{pmatrix}}}$
Problem 25

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

Notably absent from the definition of a vector space is a distance measure.

This exercise is recommended for all readers.
Problem 26
1. Prove that a sum of four vectors ${\displaystyle {\vec {v}}_{1},\ldots ,{\vec {v}}_{4}\in V}$  can be associated in any way without changing the result.
${\displaystyle {\begin{array}{rl}(({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+{\vec {v}}_{4}&=({\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3}))+{\vec {v}}_{4}\\&=({\vec {v}}_{1}+{\vec {v}}_{2})+({\vec {v}}_{3}+{\vec {v}}_{4})\\&={\vec {v}}_{1}+(({\vec {v}}_{2}+{\vec {v}}_{3})+{\vec {v}}_{4})\\&={\vec {v}}_{1}+({\vec {v}}_{2}+({\vec {v}}_{3}+{\vec {v}}_{4}))\end{array}}}$
This allows us to simply write "${\displaystyle {\vec {v}}_{1}+{\vec {v}}_{2}+{\vec {v}}_{3}+{\vec {v}}_{4}}$ " without ambiguity.
2. 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.)
1. A small rearrangement does the trick.
${\displaystyle {\begin{array}{rl}({\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3}))+{\vec {v}}_{4}&=(({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+{\vec {v}}_{4}\\&=({\vec {v}}_{1}+{\vec {v}}_{2})+({\vec {v}}_{3}+{\vec {v}}_{4})\\&={\vec {v}}_{1}+({\vec {v}}_{2}+({\vec {v}}_{3}+{\vec {v}}_{4}))\\&={\vec {v}}_{1}+(({\vec {v}}_{2}+{\vec {v}}_{3})+{\vec {v}}_{4})\end{array}}}$
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 "${\displaystyle =}$ " applies the rule ${\displaystyle ({\vec {w}}_{1}+{\vec {w}}_{2})+{\vec {w}}_{3}={\vec {w}}_{1}+({\vec {w}}_{2}+{\vec {w}}_{3})}$  by taking ${\displaystyle {\vec {w}}_{1}}$  to be ${\displaystyle {\vec {v}}_{1}+{\vec {v}}_{2}}$ , taking ${\displaystyle {\vec {w}}_{2}}$  to be ${\displaystyle {\vec {v}}_{3}}$ , and taking ${\displaystyle {\vec {w}}_{3}}$  to be ${\displaystyle {\vec {v}}_{4}}$ .
2. The base case for induction is the three vector case. This case ${\displaystyle {\vec {v}}_{1}+({\vec {v}}_{2}+{\vec {v}}_{3})=({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3}}$  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 ${\displaystyle k}$  vectors are equal no matter how the sums are parenthesized. We will show that any sum of ${\displaystyle k+1}$  vectors equals this one ${\displaystyle ((\cdots (({\vec {v}}_{1}+{\vec {v}}_{2})+{\vec {v}}_{3})+\cdots )+{\vec {v}}_{k})+{\vec {v}}_{k+1}}$ . Any parenthesized sum has an outermost "${\displaystyle +}$ ". Assume that it lies between ${\displaystyle {\vec {v}}_{m}}$  and ${\displaystyle {\vec {v}}_{m+1}}$  so the sum looks like this.
${\displaystyle (\cdots \,({\vec {v}}_{1}+(\cdots +{\vec {v}}_{m}))\,\cdots )+(\cdots \,({\vec {v}}_{m+1}+(\cdots +{\vec {v}}_{k+1}))\,\cdots )}$
The second half involves fewer than ${\displaystyle k+1}$  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 "${\displaystyle +}$ " occurs right before ${\displaystyle {\vec {v}}_{k+1}}$ .
${\displaystyle =(\cdots \,({\vec {v}}_{1}+(\,\cdots \,+{\vec {v}}_{m}))\,\cdots )+(\cdots ({\vec {v}}_{m+1}+(\cdots +{\vec {v}}_{k})\cdots )+{\vec {v}}_{k+1})}$
Apply the associativity of the sum of three things
${\displaystyle =((\,\cdots \,({\vec {v}}_{1}+(\cdots +{\vec {v}}_{m})\,\cdots \,)+(\,\cdots \,({\vec {v}}_{m+1}+(\cdots \,{\vec {v}}_{k}))\cdots )+{\vec {v}}_{k+1}}$
and finish by applying the inductive hypothesis inside these outermost parenthesis.
3. 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 ${\displaystyle \mathbb {R} ^{3}}$ ) is a subspace.

1. Show that ${\displaystyle \{a_{0}+a_{1}x+a_{2}x^{2}\,{\big |}\,a_{0}+a_{1}+a_{2}=0\}}$  is a subspace of the vector space of degree two polynomials.
2. Show that this is a subspace of the ${\displaystyle 2\!\times \!2}$  matrices.
${\displaystyle \{{\begin{pmatrix}a&b\\c&0\end{pmatrix}}\,{\big |}\,a+b=0\}}$
3. Show that a nonempty subset ${\displaystyle S}$  of a real vector space is a subspace if and only if it is closed under linear combinations of pairs of vectors: whenever ${\displaystyle c_{1},c_{2}\in \mathbb {R} }$  and ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$  then the combination ${\displaystyle c_{1}{\vec {v}}_{1}+c_{2}{\vec {v}}_{2}}$  is in ${\displaystyle S}$ .
1. We outline the check of the conditions from Definition 1.1. Additive closure holds because if ${\displaystyle a_{0}+a_{1}+a_{2}=0}$  and ${\displaystyle b_{0}+b_{1}+b_{2}=0}$  then
${\displaystyle (a_{0}+a_{1}x+a_{2}x^{2})+(b_{0}+b_{1}x+b_{2}x^{2})=(a_{0}+b_{0})+(a_{1}+b_{1})x+(a_{2}+b_{2})x^{2}}$
is in the set since ${\displaystyle (a_{0}+b_{0})+(a_{1}+b_{1})+(a_{2}+b_{2})=(a_{0}+a_{1}+a_{2})+(b_{0}+b_{1}+b_{2})}$  is zero. The second through fifth conditions are easy. Closure under scalar multiplication holds because if ${\displaystyle a_{0}+a_{1}+a_{2}=0}$  then
${\displaystyle r\cdot (a_{0}+a_{1}x+a_{2}x^{2})=(ra_{0})+(ra_{1})x+(ra_{2})x^{2}}$
is in the set as ${\displaystyle ra_{0}+ra_{1}+ra_{2}=r(a_{0}+a_{1}+a_{2})}$  is zero. The remaining conditions here are also easy.
3. Call the vector space ${\displaystyle V}$ . We have two implications: left to right, if ${\displaystyle S}$  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 ${\displaystyle S}$  is nonempty and closed, and checking the conditions of Definition 1.1. First, to show closure under addition, if ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$  then ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}\in S}$  as ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}=1\cdot {\vec {s}}_{1}+1\cdot {\vec {s}}_{2}}$ . Second, for any ${\displaystyle {\vec {s}}_{1},{\vec {s}}_{2}\in S}$ , because addition is inherited from ${\displaystyle V}$ , the sum ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}}$  in ${\displaystyle S}$  equals the sum ${\displaystyle {\vec {s}}_{1}+{\vec {s}}_{2}}$  in ${\displaystyle V}$  and that equals the sum ${\displaystyle {\vec {s}}_{2}+{\vec {s}}_{1}}$  in ${\displaystyle V}$  and that in turn equals the sum ${\displaystyle {\vec {s}}_{2}+{\vec {s}}_{1}}$  in ${\displaystyle S}$ . The argument for the third condition is similar to that for the second. For the fourth, suppose that ${\displaystyle {\vec {s}}}$  is in the nonempty set ${\displaystyle S}$  and note that ${\displaystyle 0\cdot {\vec {s}}={\vec {0}}\in S}$ ; showing that the ${\displaystyle {\vec {0}}}$  of ${\displaystyle V}$  acts under the inherited operations as the additive identity of ${\displaystyle S}$  is easy. The fifth condition is satisfied because for any ${\displaystyle {\vec {s}}\in S}$  closure under linear combinations shows that the vector ${\displaystyle 0\cdot {\vec {0}}+(-1)\cdot {\vec {s}}}$  is in ${\displaystyle S}$ ; showing that it is the additive inverse of ${\displaystyle {\vec {s}}}$  under the inherited operations is routine. The proofs for the remaining conditions are similar.