Education: Solutions

SOLUTIONS TO EXERCISES-Vector Spaces ( definitions and examples):

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 $\mathcal{P}_1=\{a_0+a_1x\,\big|\, a_0,a_1\in\mathbb{R}\}$ 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
$L=\{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\in\mathbb{R}^4\,\big|\, x+y-z+w=0\}$
under the operations inherited from $\mathbb{R}^4$ .

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
$\begin{pmatrix} x_1 \\ y_1 \\ z_1 \\ w_1 \end{pmatrix} +\begin{pmatrix} x_2 \\ y_2 \\ z_2 \\ w_2 \end{pmatrix} = \begin{pmatrix} x_1+x_2 \\ y_1+y_2 \\ z_1+z_2 \\ w_1+w_2 \end{pmatrix}$
is in $L$ 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.)

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,
$\{a_0+a_1x+a_2x^2\,\big|\, a_0,a_1,a_2\in\mathbb{R}^+\}$
where $\mathbb{R}^+$ is the set of reals greater than zero
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_1i)+(w_0+w_1i)=(v_0+w_0)+(v_1+w_1)i

and

r(v_0+v_1i)=(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_1x+v_2y+v_3z)+(w_1x+w_2y+w_3z)=(v_1+w_1)x+(v_2+w_2)y+(v_3+w_3)z

and

r\cdot(v_1x+v_2y+v_3z)=(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
$\{\begin{pmatrix} x \\ y \\ z \\ w \end{pmatrix}\in\mathbb{R}^4 \,\big|\, x+y+w=1\}$
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

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^rw^r

. The ninth condition asserts that

(v^r)^s=v^{rs}

. The final condition says that

v^1=v

Problem 13

\{(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 contition 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
$\{\begin{pmatrix} a &0 \\ b &-a-b \end{pmatrix} \,\big|\, a,b\in\mathbb{R}\}$
so that there are two parameters.

This exercise is recommended for all readers.

Problem 16

Using

\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 $(1/r)\cdot r\vec{v}=(1/r)\cdot \vec{0}$ 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_1i)+(b_0+b_1i)=(a_0+b_0)+(a_1+b_1)i

and

(a_0+a_1i)(b_0+b_1i)=(a_0b_0-a_1b_1)+(a_0b_1+a_1b_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.
$\begin{array}{rl} ((\vec{v}_1+\vec{v}_2)+\vec{v}_3)+\vec{v}_4 &=(\vec{v}_1+(\vec{v}_2+\vec{v}_3))+\vec{v}_4 \\ &=(\vec{v}_1+\vec{v}_2)+(\vec{v}_3+\vec{v}_4) \\ &=\vec{v}_1+((\vec{v}_2+\vec{v}_3)+\vec{v}_4) \\ &=\vec{v}_1+(\vec{v}_2+(\vec{v}_3+\vec{v}_4)) \end{array}$
This allows us to simply write " $\vec{v}_1+\vec{v}_2+\vec{v}_3+\vec{v}_4$ " without ambiguity.
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.
  $\begin{array}{rl} (\vec{v}_1+(\vec{v}_2+\vec{v}_3))+\vec{v}_4 &=((\vec{v}_1+\vec{v}_2)+\vec{v}_3)+\vec{v}_4 \\ &=(\vec{v}_1+\vec{v}_2)+(\vec{v}_3+\vec{v}_4) \\ &=\vec{v}_1+(\vec{v}_2+(\vec{v}_3+\vec{v}_4)) \\ &=\vec{v}_1+((\vec{v}_2+\vec{v}_3)+\vec{v}_4) \end{array}$
  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 " $=$ " 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 $\vec{v}_4$ .
2. 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.
  $(\cdots\,(\vec{v}_1+(\cdots+\vec{v}_m))\,\cdots)+(\cdots\,(\vec{v}_{m+1}+(\cdots+\vec{v}_{k+1}))\,\cdots)$
  The second half involves fewer than $k+1$ 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 $\vec{v}_{k+1}$ .
  $=(\cdots\,(\vec{v}_1+(\,\cdots\,+\vec{v}_m))\,\cdots)+(\cdots(\vec{v}_{m+1}+(\cdots+\vec{v}_{k})\cdots)+\vec{v}_{k+1})$
  Apply the associativity of the sum of three things
  $=((\,\cdots\,(\vec{v}_1+(\cdots+\vec{v}_m)\,\cdots\,) +(\,\cdots\,(\vec{v}_{m+1}+(\cdots\,\vec{v}_k))\cdots) +\vec{v}_{k+1}$
  and finish by applying the inductive hypothesis inside these outermost parenthesis.

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 $\{a_0+a_1x+a_2x^2\,\big|\, a_0+a_1+a_2=0\}$ 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
$(a_0+a_1x+a_2x^2)+(b_0+b_1x+b_2x^2)= (a_0+b_0)+(a_1+b_1)x+(a_2+b_2)x^2$
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
$r\cdot(a_0+a_1x+a_2x^2)= (ra_0)+(ra_1)x+(ra_2)x^2$
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.
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

15/09/2014

Solutions