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 $2\times 4$ matrices
3. The space $\{f:[0,1]\to ℝ|f\text{ is continuous}\}$
4. The space of real-valued functions of one natural number variable

Answer

1. $0+0x+0x^2+0x^3$
2. $\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)$
3. The constant function $f(x)=0$
4. 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.

1. In ${𝒫}_3$, the vector $-3-2x+x^2$.
2. In the space $2\times 2$,

   $\left(\begin{array}{cc}1& -1\\ 0& 3\end{array}\right).$

3. In $\{a{e}^x+b{e}^{-x}|a,b\in ℝ\}$, the space of functions of the real variable $x$ under the natural operations, the vector $3e^x-2e^{-x}$.

Answer

1. $3+2x-x^2$
2. $\left(\begin{array}{cc}-1& +1\\ 0& -3\end{array}\right)$
3. $-3e^x+2e^{-x}$

This exercise is recommended for all readers.

Problem 3

Show that each of these is a vector space.

1. The set of linear polynomials ${𝒫}_1=\{a_0+a_{1}x|a_0,a_1\in ℝ\}$ under the usual polynomial addition and scalar multiplication operations.
2. The set of $2\times 2$ matrices with real entries under the usual matrix operations.
3. The set of three-component row vectors with their usual operations.
4. The set

   $L=\{\left(\begin{array}{c}x\\ y\\ z\\ w\end{array}\right)\in {ℝ}^4|x+y-z+w=0\}$

   under the operations inherited from ${ℝ}^4$.

Answer

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 $0+0x$.
2. The zero element of this space is the $2\times 2$ matrix of zeroes.
3. The zero element is the vector of zeroes.
4. Closure of addition involves noting that the sum

   $\left(\begin{array}{c}{x}_1\\ y_1\\ z_1\\ w_1\end{array}\right)+\left(\begin{array}{c}{x}_2\\ y_2\\ z_2\\ w_2\end{array}\right)=\left(\begin{array}{c}{x}_1+x_2\\ y_1+y_2\\ z_1+z_2\\ w_1+w_2\end{array}\right)$

   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 $L$.

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 ${ℝ}^3$, this set

   $\{\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\in {ℝ}^3|x+y+z=1\}$

2. Under the operations inherited from ${ℝ}^3$, this set

   $\{\left(\begin{array}{c}x\\ y\\ z\end{array}\right)\in {ℝ}^3|x^2+y^2+z^2=1\}$

3. Under the usual matrix operations,

   $\{\left(\begin{array}{cc}a& 1\\ b& c\end{array}\right)|a,b,c\in ℝ\}$

4. Under the usual polynomial operations,

   $\{a_0+a_{1}x+a_2{x}^2|a_0,a_1,a_2\in {ℝ}^{+}\}$

   where ${ℝ}^{+}$ is the set of reals greater than zero
5. Under the inherited operations,

   $\{\left(\begin{array}{c}x\\ y\end{array}\right)\in {ℝ}^2|x+3y=4\text{ and }2x-y=3\text{ and }6x+4y=10\}$

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.

1. It is not closed under addition; it fails to meet condition 1.

   $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\in Q\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\in ̸Q$

2. It is not closed under addition.

   $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)\in Q\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right)\in ̸Q$

3. It is not closed under addition.

   $\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\in Q\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)\in ̸Q$

4. It is not closed under scalar multiplication.

   $1+1x+1x^2\in Q-1\cdot (1+1x+1x^2)\in ̸Q$

5. It is empty, violating condition 4.

Problem 5

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

Answer

The usual operations $(v_0+v_{1}i)+(w_0+w_{1}i)=(v_0+w_0)+(v_1+w_1)i$ and $r(v_0+v_{1}i)=(r{v}_0)+(r{v}_1)i$ suffice. The check is easy.

This exercise is recommended for all readers.

Problem 6

Is the set of rational numbers a vector space over $ℝ$ 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 $r\cdot (v_{1}x+v_{2}y+v_{3}z)=(r{v}_1)x+(r{v}_2)y+(r{v}_3)z$. 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.

$\left(\begin{array}{c}{x}_1\\ y_1\end{array}\right)+\left(\begin{array}{c}{x}_2\\ y_2\end{array}\right)=\left(\begin{array}{c}{x}_1-x_2\\ y_1-y_2\end{array}\right)r\cdot \left(\begin{array}{c}x\\ y\end{array}\right)=\left(\begin{array}{c}rx\\ ry\end{array}\right)$

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 ${ℝ}^3$ is a vector space under these operations.

1. $\left(\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right)+\left(\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\text{and}r\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}rx\\ ry\\ rz\end{array}\right)$
2. $\left(\begin{array}{c}{x}_1\\ y_1\\ z_1\end{array}\right)+\left(\begin{array}{c}{x}_2\\ y_2\\ z_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\text{and}r\left(\begin{array}{c}x\\ y\\ z\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)$

Answer

1. It is not a vector space.

   $(1+1)\cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\ne \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$

2. It is not a vector space.

   $1\cdot \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)\ne \left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$

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 $2\times 2$ matrices

   $\{\left(\begin{array}{cc}a& 0\\ 0& b\end{array}\right)|a,b\in ℝ\}$

2. This set of $2\times 2$ matrices

   $\{\left(\begin{array}{cc}x& x+y\\ x+y& y\end{array}\right)|x,y\in ℝ\}$

3. This set

   $\{\left(\begin{array}{c}x\\ y\\ z\\ w\end{array}\right)\in {ℝ}^4|x+y+z+w=1\}$

4. The set of functions $\{f:ℝ\to ℝ|df/dx+2f=0\}$
5. The set of functions $\{f:ℝ\to ℝ|df/dx+2f=1\}$

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.

1. Yes.
2. Yes.
3. No, it is not closed under addition. The vector of all $1/4$'s, when added to itself, makes a nonmember.
4. Yes.
5. 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 ${ℝ}^{+}$ 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 $1\in {ℝ}^{+}$ 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 $\{(x,y)|x,y\in ℝ\}$ a vector space under these operations?

1. $(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$ and $r\cdot (x,y)=(rx,y)$
2. $(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$ and $r\cdot (x,y)=(rx,0)$

Answer

1. No: $1\cdot (0,1)+1\cdot (0,1)\ne (1+1)\cdot (0,1)$.
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 $1\cdot (0,1)\ne (0,1)$.

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 ${ℝ}^k$.

1. The $2\times 3$ matrices under the usual operations
2. The $n\times m$ matrices (under their usual operations)
3. This set of $2\times 2$ matrices

   $\{\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)|a,b,c\in ℝ\}$

4. This set of $2\times 2$ matrices

   $\{\left(\begin{array}{cc}a& 0\\ b& c\end{array}\right)|a+b+c=0\}$

Answer

1. $6$
2. $nm$
3. $3$
4. To see that the answer is $2$, rewrite it as

   $\{\left(\begin{array}{cc}a& 0\\ b& -a-b\end{array}\right)|a,b\in ℝ\}$

   so that there are two parameters.

This exercise is recommended for all readers.

Problem 16

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

Answer

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

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

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 $\overrightarrow{v},\overrightarrow{s},\overrightarrow{t}\in V$ then $\overrightarrow{v}+\overrightarrow{s}=\overrightarrow{v}+\overrightarrow{t}$ implies that $\overrightarrow{s}=\overrightarrow{t}$.

Answer

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

Problem 18

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

Answer

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

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 $\overrightarrow{v}\in V$. If ${\overrightarrow{w}}_1,{\overrightarrow{w}}_2\in V$ are both additive inverses of $\overrightarrow{v}$ then consider ${\overrightarrow{w}}_1+\overrightarrow{v}+{\overrightarrow{w}}_2$. On the one hand, we have that it equals ${\overrightarrow{w}}_1+(\overrightarrow{v}+{\overrightarrow{w}}_2)={\overrightarrow{w}}_1+\overrightarrow{0}={\overrightarrow{w}}_1$. On the other hand we have that it equals $({\overrightarrow{w}}_1+\overrightarrow{v})+{\overrightarrow{w}}_2=\overrightarrow{0}+{\overrightarrow{w}}_2={\overrightarrow{w}}_2$. Therefore, ${\overrightarrow{w}}_1={\overrightarrow{w}}_2$.

Problem 21

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

Answer

1. Every such set has the form $\{r\cdot \overrightarrow{v}+s\cdot \overrightarrow{w}|r,s\in ℝ\}$ where either or both of $\overrightarrow{v},\overrightarrow{w}$ may be $\overrightarrow{0}$. With the inherited operations, closure of addition $(r_1\overrightarrow{v}+s_1\overrightarrow{w})+(r_2\overrightarrow{v}+s_2\overrightarrow{w})=(r_1+r_2)\overrightarrow{v}+(s_1+s_2)\overrightarrow{w}$ and scalar multiplication $c(r\overrightarrow{v}+s\overrightarrow{w})=(cr)\overrightarrow{v}+(cs)\overrightarrow{w}$ 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 $\overrightarrow{v}\in V$ is not $\overrightarrow{0}$.

1. Prove that $r\cdot \overrightarrow{v}=\overrightarrow{0}$ if and only if $r=0$.
2. Prove that $r_1\cdot \overrightarrow{v}=r_2\cdot \overrightarrow{v}$ if and only if $r_1=r_2$.
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.

Answer

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

1. One direction of the if and only if is clear: if $r=0$ then $r\cdot \overrightarrow{v}=\overrightarrow{0}$. For the other way, let $r$ be a nonzero scalar. If $r\overrightarrow{v}=\overrightarrow{0}$ then $1\cdot \overrightarrow{v}=(r/r)\cdot \overrightarrow{v}=(1/r)\cdot r\overrightarrow{v}=(1/r)\cdot \overrightarrow{0}$ shows that $\overrightarrow{v}=\overrightarrow{0}$, contrary to the assumption.
2. Where $r_1,r_2$ are scalars, $r_1\overrightarrow{v}=r_2\overrightarrow{v}$ holds if and only if $(r_1-r_2)\overrightarrow{v}=\overrightarrow{0}$. By the prior item, then $r_1-r_2=0$.
3. A nontrivial space has a vector $\overrightarrow{v}\ne \overrightarrow{0}$. Consider the set $\{k\cdot \overrightarrow{v}|k\in ℝ\}$. 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?

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 $ℂ$ has the same definition as a vector space over the reals except that scalars are drawn from $ℂ$ instead of from $ℝ$. 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 $(a_0+a_{1}i)(b_0+b_{1}i)=(a_0{b}_0-a_1{b}_1)+(a_0{b}_1+a_1{b}_0)i$.)

1. The set of degree two polynomials with complex coefficients
2. This set

   $\{\left(\begin{array}{cc}0& a\\ b& 0\end{array}\right)|a,b\in ℂ\text{ and }a+b=0+0i\}$

Answer

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

1. $(0+0i)+(0+0i)x+(0+0i)x^2$
2. $\left(\begin{array}{cc}0+0i& 0+0i\\ 0+0i& 0+0i\end{array}\right)$

Problem 25

Name a property shared by all of the ${ℝ}^n$'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

1. Prove that a sum of four vectors ${\overrightarrow{v}}_1,\dots ,{\overrightarrow{v}}_4\in V$ can be associated in any way without changing the result.

   $\begin{array}{cc}(({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)+{\overrightarrow{v}}_3)+{\overrightarrow{v}}_4& =({\overrightarrow{v}}_1+({\overrightarrow{v}}_2+{\overrightarrow{v}}_3))+{\overrightarrow{v}}_4\\ & =({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)+({\overrightarrow{v}}_3+{\overrightarrow{v}}_4)\\ & ={\overrightarrow{v}}_1+(({\overrightarrow{v}}_2+{\overrightarrow{v}}_3)+{\overrightarrow{v}}_4)\\ & ={\overrightarrow{v}}_1+({\overrightarrow{v}}_2+({\overrightarrow{v}}_3+{\overrightarrow{v}}_4))\end{array}$

   This allows us to simply write "${\overrightarrow{v}}_1+{\overrightarrow{v}}_2+{\overrightarrow{v}}_3+{\overrightarrow{v}}_4$" 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.)

Answer

1. A small rearrangement does the trick.

   $\begin{array}{cc}({\overrightarrow{v}}_1+({\overrightarrow{v}}_2+{\overrightarrow{v}}_3))+{\overrightarrow{v}}_4& =(({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)+{\overrightarrow{v}}_3)+{\overrightarrow{v}}_4\\ & =({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)+({\overrightarrow{v}}_3+{\overrightarrow{v}}_4)\\ & ={\overrightarrow{v}}_1+({\overrightarrow{v}}_2+({\overrightarrow{v}}_3+{\overrightarrow{v}}_4))\\ & ={\overrightarrow{v}}_1+(({\overrightarrow{v}}_2+{\overrightarrow{v}}_3)+{\overrightarrow{v}}_4)\end{array}$

   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 "$=$" applies the rule $({\overrightarrow{w}}_1+{\overrightarrow{w}}_2)+{\overrightarrow{w}}_3={\overrightarrow{w}}_1+({\overrightarrow{w}}_2+{\overrightarrow{w}}_3)$ by taking ${\overrightarrow{w}}_1$ to be ${\overrightarrow{v}}_1+{\overrightarrow{v}}_2$, taking ${\overrightarrow{w}}_2$ to be ${\overrightarrow{v}}_3$, and taking ${\overrightarrow{w}}_3$ to be ${\overrightarrow{v}}_4$.
2. The base case for induction is the three vector case. This case ${\overrightarrow{v}}_1+({\overrightarrow{v}}_2+{\overrightarrow{v}}_3)=({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)+{\overrightarrow{v}}_3$ 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 $((\cdots (({\overrightarrow{v}}_1+{\overrightarrow{v}}_2)+{\overrightarrow{v}}_3)+\cdots )+{\overrightarrow{v}}_k)+{\overrightarrow{v}}_{k+1}$. Any parenthesized sum has an outermost "$+$". Assume that it lies between ${\overrightarrow{v}}_m$ and ${\overrightarrow{v}}_{m+1}$ so the sum looks like this.

   $(\cdots ({\overrightarrow{v}}_1+(\cdots +{\overrightarrow{v}}_m))\cdots )+(\cdots ({\overrightarrow{v}}_{m+1}+(\cdots +{\overrightarrow{v}}_{k+1}))\cdots )$

   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 ${\overrightarrow{v}}_{k+1}$.

   $=(\cdots ({\overrightarrow{v}}_1+(\cdots +{\overrightarrow{v}}_m))\cdots )+(\cdots ({\overrightarrow{v}}_{m+1}+(\cdots +{\overrightarrow{v}}_k)\cdots )+{\overrightarrow{v}}_{k+1})$

   Apply the associativity of the sum of three things

   $=((\cdots ({\overrightarrow{v}}_1+(\cdots +{\overrightarrow{v}}_m)\cdots )+(\cdots ({\overrightarrow{v}}_{m+1}+(\cdots {\overrightarrow{v}}_k))\cdots )+{\overrightarrow{v}}_{k+1}$

   and finish by applying the inductive hypothesis inside these outermost parenthesis.

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 ${ℝ}^3$) is a subspace.

1. Show that $\{a_0+a_{1}x+a_2{x}^2|a_0+a_1+a_2=0\}$ is a subspace of the vector space of degree two polynomials.
2. Show that this is a subspace of the $2\times 2$ matrices.

   $\{\left(\begin{array}{cc}a& b\\ c& 0\end{array}\right)|a+b=0\}$

3. 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 $c_1,c_2\in ℝ$ and ${\overrightarrow{s}}_1,{\overrightarrow{s}}_2\in S$ then the combination $c_1{\overrightarrow{v}}_1+c_2{\overrightarrow{v}}_2$ is in $S$.

Answer

1. 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

   $(a_0+a_{1}x+a_2{x}^2)+(b_0+b_{1}x+b_2{x}^2)=(a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2$

   is in the set since $(a_0+b_0)+(a_1+b_1)+(a_2+b_2)=(a_0+a_1+a_2)+(b_0+b_1+b_2)$ is zero. The second through fifth conditions are easy. Closure under scalar multiplication holds because if $a_0+a_1+a_2=0$ then

   $r\cdot (a_0+a_{1}x+a_2{x}^2)=(r{a}_0)+(r{a}_1)x+(r{a}_2)x^2$

   is in the set as $r{a}_0+r{a}_1+r{a}_2=r(a_0+a_1+a_2)$ is zero. The remaining conditions here are also easy.
2. This is similar to the prior answer.
3. 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 ${\overrightarrow{s}}_1,{\overrightarrow{s}}_2\in S$ then ${\overrightarrow{s}}_1+{\overrightarrow{s}}_2\in S$ as ${\overrightarrow{s}}_1+{\overrightarrow{s}}_2=1\cdot {\overrightarrow{s}}_1+1\cdot {\overrightarrow{s}}_2$. Second, for any ${\overrightarrow{s}}_1,{\overrightarrow{s}}_2\in S$, because addition is inherited from $V$, the sum ${\overrightarrow{s}}_1+{\overrightarrow{s}}_2$ in $S$ equals the sum ${\overrightarrow{s}}_1+{\overrightarrow{s}}_2$ in $V$ and that equals the sum ${\overrightarrow{s}}_2+{\overrightarrow{s}}_1$ in $V$ and that in turn equals the sum ${\overrightarrow{s}}_2+{\overrightarrow{s}}_1$ in $S$. The argument for the third condition is similar to that for the second. For the fourth, suppose that $\overrightarrow{s}$ is in the nonempty set $S$ and note that $0\cdot \overrightarrow{s}=\overrightarrow{0}\in S$; showing that the $\overrightarrow{0}$ of $V$ acts under the inherited operations as the additive identity of $S$ is easy. The fifth condition is satisfied because for any $\overrightarrow{s}\in S$ closure under linear combinations shows that the vector $0\cdot \overrightarrow{0}+(-1)\cdot \overrightarrow{s}$ is in $S$; showing that it is the additive inverse of $\overrightarrow{s}$ under the inherited operations is routine. The proofs for the remaining conditions are similar.