Homework 2 Solutions

Chapter 1, section 2 (page 12 in the 4th edition):

Problem 1.

These are true/false questions. For each question give a one sentence explanation of your answer.

a True

b False:

proved in lecture

c False:

if $x=0_V$ then $ax=bx$ holds even if $a\neq b$

d False:

if $a=0_{\mathbb F}$ then $ax=ay$ even if $x$ and $y$ are different vectors.

e 'Trueish'

The phrase "may be regarded as" is a bit fishy, i.e. it has no precise mathematical meaning. A vector in $\mathbb{F}^n$ is an $n$-tuple of numbers $(x_1, \dots, x_n)$ and an $n\times1$ matrix is a row vector $[x_1\; x_2\; \cdots \; x_n]$, so you could say they are the same.

Would you say that a $n$-tuple $(x_1, \dots, x_n)$ and a column vector $\begin{bmatrix} x_1 \\ \cdots \\ x_n\end{bmatrix}$ “are the same?” I am not sure what the authors of our textbook would think of this, but they would definitely insist that a row vector $[x_1 \dots x_n]$ and a column vector $\begin{bmatrix} x_1 \\ \cdots \\ x_n\end{bmatrix}$ are not the same.

There is a precise mathematical way to say “may be regarded as” which involves the concept of “isomorphism.” We will see this later in the semester.

f False

It's the other way around

g False

you can add polynomials in $\mathcal{P}(\mathbb{F})$ no matter what their degree is.

h False

The degree could be lower if terms cancel. E.g. $f(X)=1+X$ and $g(X)=1-X$ both have degree 1, but their sum $f(X)+g(X)=2$ has degree zero.

i True

j Trueish

Again, what does "may be considered" mean?

k True, by definition

Problem 2

$0_{M_{3\times 4}(\mathbb{F})} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

Problem 3

$M_{13}=3$, $M_{21}=4$, $M_{22}=5$.

Problem 7

$f$ and $g$ are functions, so, by definition, $f=g$ means $f(t)=g(t)$ for all $t\in S$. The set $S$ has only two elements and we can compute $f(t)$ and $g(t)$ for both elements of $S$ and compare:

So we see that $f(t)=g(t)$ and $f(t)+g(t)=h(t)$ for all (both) $t\in S$. Therefore $f=g$ and $h=f+g$.

Problem 8: in writing your proof, say which axioms you use

We are asked to show that $(a+b)(x+y)=ax+bx+ay+by$ for all $a,b\in\mathbb{F}$ and all $x,y\in V$.

Proof:

$\begin{aligned} (a+b)(x+y) &= (a+b)x + (a+b)y &\text{ (distributive law VS7)}\\ &= (ax + bx) + (ay + by) & \text{(distributive law VS8)}\\ &= ax + (bx + (ay + by)) &\text{(associative law VS2)}\\ &= ax + ((bx + ay) + by)) &\text{(associative law VS2)}\\ &= ax + ((ay + bx) + by)) &\text{(commutative law VS1)}\\ &= ax + (ay + (bx + by)) &\text{(associative law VS2)}\\ &= (ax + ay) + (bx + by) &\text{(associative law VS2)}\\ &= ax + ay + bx + by.& \end{aligned}$

In the last step we omited the parentheses because the associative law (VS2) says that the order in which the additions are performed does not affect the outcome.

Problem 12

$V=\mathcal{F}(\R,\R)$, $W=\{f\in V \mid f\text{ is even}\}$, show that $W$ is a vector space.

You could verify all the hypothesis of a vector space, but you could also show that $W$ is a linear subspace of $V$ and use a theorem from the next section.

To prove that $W$ is a subspace we check that it is (1) not empty, (2) closed under vector addition, and (3) closed under multiplication with scalars.

1. The zero function, defined by $f(t)=0$, is even because for all $t\in \R$ one has $f(t) = 0$ and $f(-t)=0$ so that $f(t)=f(-t)$: $W$ is not empty

2. If $f$ and $g$ are even functions then $h=f+g$ satisfies

$\begin{aligned} h(-t) &= f(-t)+g(-t) &\text{def of }h=f+g\\ &=f(t)+g(t) &\text{$f$ and $g$ are even}\\ &=h(t) &\text{def of $h=f+g$ again} \end{aligned}$

for all $t\in\R$. Therefore $h$ is even. $W$ is closed under addition.

3. If $f$ is an even function and $a\in\R$ then $g=af$ satisfies

$\begin{aligned} g(-t) &= af(-t) & \text{def of $g=af$}\\ &= af(t) &\text{$f$ is even}\\ &= g(t) &\text{def of $g=af$ again} \end{aligned}$

for every $t\in\R$. Therefore $g$ is even and therefore $W$ is closed under multiplication with scalars.

Problem 13

Given $V=\{(x_1,x_2) \mid x_1, x_2\in\R\}$, $\mathbb{F}=\R$, and define addition and scalar multiplication by

$(a_1,a_2)+(b_1,b_2) = (a_1+b_1, a_2b_2), \qquad c(a_1, a_2) = (ca_1, a_2).$

To distinguish these rules from the usual addition and scalar multiplication in $\R^2$ let's refer to them as hokey addition and hokey multiplication.

Question: Does this define a vector space?

Answer: No. Several of the axioms fail. While we only need one failing axiom to show that $(V,\mathbb{F})$ is not a vector space, let's go through the list to see which ones fail and which ones don't.

VS1, VS2 hokey addition is commutative and associative.

VS3 The element $z=(0,1)$ acts as a zero vector for hokey addition: for all $a=(a_1,a_2)\in V$ one has $a + z = (a_1,a_2)+(0,1) = (a_1+0, a_2\cdot1) = (a_1, a_2)$. If there were another zero vector, say $\tilde z$, then VS1–3 imply $z=z+\tilde z = \tilde z$. Therefore there is only one zero vector (we don't need the axioms VS4–8 for this.)

VS4 This axiom fails because the zero vector is $z=(0,1)$, as we proved above. If $x=(0,0)$ then $x\neq z$ so that $x$ is not the zero vector for hokey addition in $V$. If axiom VS4 were true then there would be a $y=(y_1, y_2)\in V$ such that $x+y=z$. But by the definition of hokey addition we have $x+y = (0,0)+(y_1,y_2) = (y_1, 0)\neq (0,1)$. So axiom VS4 fails.

VS5 This holds: $1\cdot(a_1,a_2) = (1\cdot a_1, a_2) = (a_1,a_2)$.

VS6 Also holds

VS7 This holds. To prove that VS7 actually holds you compute

$a\bigl((x_1,x_2)+(y_1,y_2)\bigr) = a(x_1+y_1, x_2y_2)=(ax_1+ay_1, x_2y_2)$

and

$a(x_1,x_2)+a(y_1,y_2) = (ax_1,x_2)+(ay_1,y_2) = (ax_1+ay_1, x_2y_2)$

Therefore

$a\bigl((x_1,x_2)+(y_1,y_2)\bigr) = a(x_1,x_2)+a(y_1,y_2)$

VS8 This fails. When you compute $(a+b)x$ and $ax+bx$ for $x=(x_1,x_2)$ using the rules for hokey addition and multiplication you find

$(a+b)(x_1,x_2) = \bigl((a+b)x_1, x_2\bigr) = (ax_1+bx_1, x_2)$

and on the other hand

$a(x_1,x_2)+b(x_1,x_2) = (ax_1, x_2)+ (bx_1, x_2) = (ax_1+bx_1, (x_2)^2)$

These are different as long as $x_2\neq0$ and $x_2\neq 1$ (so that $x_2\neq (x_2)^2$).

Problem 14

$V=\mathbb{C}^n$ is a vectorspace over $\mathbb{F}=\mathbb{C}$. Is it a vector space if we replace the field of numbers by $\R$? More precisely, suppose we set $V=\mathbb{C}^n$ and $\mathbb{F}=\R$ and define addition and multiplication by

$(x_1,\dots,x_n)+(y_1,\dots,y_n) = (x_1+y_1, \dots, x_n+y_n)$

and

$a(x_1,\dots, x_n) = (ax_1, \dots, ax_n).$

Is $(V, \mathbb{F})$ then a vector space?

Answer: yes.

Reason: vector addition and scalar multiplication are exactly the same as those for $(\mathbb{C}^n, \mathbb{C})$ so the axioms VS–8 still hold.

Problem 15

A very similar problem, but not exactly the same:

$V=\mathbb{R}^n$ is a vectorspace over $\mathbb{F}=\mathbb{R}$. Is it a vector space if we replace the field of numbers by $\mathbb{C}$? More precisely, suppose we set $V=\mathbb{R}^n$ and $\mathbb{F}=\mathbb{C}$ and define addition and multiplication by

$(x_1,\dots,x_n)+(y_1,\dots,y_n) = (x_1+y_1, \dots, x_n+y_n)$

and

$a(x_1,\dots, x_n) = (ax_1, \dots, ax_n).$

Is $(V, \mathbb{F})$ then a vector space?

Answer: The definition of multiplication is broken because $ax$ is not always a vector in $\R^n$. For example, if $a=i=\sqrt{-1}$ and $x=(1,1)$ then the definition of multiplication would make us say $ax = (i,i)$. But $i$ is not a real number so $(i,i)\not\in\R^n$. Since multiplication is not well defined we cannot verify the axioms, and this $(V,\mathbb{F})$ is not a vector space.