Homework 4 Solutions

§2.1

1—true/false questions

a. If $T$ is linear, then $T$ preserves sums and scalar products. True. The phrase “$T$ preserves sums” means $T(x+y)=Tx+Ty$ for all $x, y$.

b. If $T(x + y) = T(x) + T(y)$, then $T$ is linear. False. A counterexample is difficult to describe, but you do need to check $T(cx)=cTx$ for all $x\in V$, $c\in\mathbb F$.

c. $T$ is one-to-one if and only if the only vector $x$ such that $T(x) = 0$ is $x = 0$ . True if $T$ is linear. This is the injectivity theorem. On the other hand this is a bit of a "Simon says" problem, in that the problem does not say that $T$ is linear. Since $T$ is allowed to be nonlinear the answer is false. Examples:

• $T(x)=x-1$ is a function from $\R\to\R$ that is injective, but $\{x\mid T(x)=0\} = \{1\}\neq\{0\}$.
• $T(x)=x^2$ is a function for which $\{x\in\R\mid T(x)=0\}=\{0\}$, but $T$ is not injective (e.g. $T(-1)=T(+1)$).

So neither implication in “$T$ injective”$\iff \{x\mid T(x)=0\}=\{0\}$ is true.

d. If $T$ is linear, then $T(0_V ) = 0_W$. True.

e. If $T:V\to W$ is linear, then $\mathrm{nullity}(T) + \mathrm{rank}(T) = \dim(W)$.
False. This looks like the rank+nullity theorem, but $\dim W$ actually does not appear in the theorem. It really says $\mathrm{nullity}(T) + \mathrm{rank}(T) = \dim(V)$.

f. If $T$ is linear, then $T$ carries linearly independent subsets of $V$ onto linearly independent subsets of $W$. False. This statement claims that if $\{v_1, \dots, v_k\}\subset V$ is linearly independent, then $\{Tv_1, \dots , Tv_k\}\subset W$ also is independent. This is not true. For example, if one or more of the vectors $v_k$ belong to the null space of $T$ then the set of vectors $\{Tv_1, \dots , Tv_k\}$ contains the zero vector and therefore is not independent.

g. If $T, S : V → W$ are both linear and agree on a basis for $V$, then $T = S$. True. We are given two linear maps $T, S:V\to W$, and a basis $\{v_1, \dots, v_k\}$ of $V$ such that $Tv_1=Sv_1$, … , $Tv_k=Sv_k$. If $x\in V$ is any vector, then there exist $x_1, \dots, x_k\in \mathbb F$ such that $x=x_1v_1+\cdots+x_kv_k$. Then we have

$\begin{aligned} Tx&=T\left(x_1v_1+\cdots+x_kv_k\right)\\ &=x_1Tv_1+\cdots+x_kTv_k \\ &=x_1Sv_1+\cdots+x_kSv_k \\ &=S\left(x_1v_1+\cdots+x_kv_k\right)\\ &= Sx. \end{aligned}$

Therefore $T$ and $S$ are the same linear transformation.

2

To show that $T(a_1 , a_2 , a_3 ) = (a_1 − a_2 , 2a_3 )$ defines a linear transformation, verify $T(a+b)=Ta+Tb$ and $T(ta)=t\, Ta$ for all $a,b\in\mathbb F^3$ and $t\in \mathbb F$.

The Null Space. By definition $a\in N(T)\iff Ta=0$, so to find all $a\in N(T)$ we solve $Ta=0$, i.e.

$Ta=T\begin{pmatrix} a_1 \\ a_2\\ a_3 \end{pmatrix} = \binom{a_1-a_2}{2a_3} = \binom00 \iff a_1=a_2 \text{ and }a_3=0.$

Thus $N(T)$ consists of all vectors of the form

$a=\begin{pmatrix} a_1 \\a_1 \\0 \end{pmatrix} =a_1 \begin{pmatrix} 1\\1\\0 \end{pmatrix}$

In set notation:

$N(T) = \left\{a_1 \begin{pmatrix} 1\\1\\0 \end{pmatrix} \Bigg| a_1\in\R\right\}.$

This shows that $N(T)$ is one dimensional, and $\left(\begin{smallmatrix} 1 \\ 1\\ 0 \end{smallmatrix}\right)$ is a basis for $N(T)$.

The Range. The nullity+rank theorem says $\dim N(T)+\dim R(T)=\dim \R^3$. Since we have found that $\dim N(T)=1$ we have $\dim R(T)=3-1=2$: the range of $T$ is two dimensional; the rank of $T$ is 2.

Since $R(T)\subset\R^2$ and $\R^2$ is also two dimensional, we conclude that $R(T)=\R^2$. $T$ is “onto,” or surjective.

5

We consider $T:\mathcal{P}_2(\R)\to \mathcal{P}_3(\R)$ given by $Tf(x) = xf(x)+f'(x)$.

Linearity. For any $f, g\in \mathcal P_2(\R)$ one has

$\begin{aligned} T(f+g)(x) &= x(f+g)(x) + (f+g)'(x) \\ &= x\left(f(x)+g(x)\right) + f'(x) + g'(x) \\ &= xf(x) + f'(x) + xg(x)+g'(x)\\ &= (Tf)(x) + (Tg)(x) \end{aligned}$

Hence $T(f+g) = Tf + Tg$ for any two polynomials $f,g\in \mathcal P_2(\R)$.

One verifies $T(af) = aTf$ with a similar computation.

This shows $T$ is linear.

The Null Space. By definition $f\in N(T)$ if $f\in \mathcal{P}_2(\R)$ and $Tf(x) = xf(x)+f'(x)=0$. This looks like a differential equation, and if you remembered the right calculus you could try solving the diffeq. But because of the extra assumption that $f$ is a polynomial this is an algebra problem and we don't have to use any calculus beyond the definition of $f'(x)$.

If $f\in\mathcal{P}_2(\R)$ then $f(x) = a+bx+cx^2$ for certain $a,b,c\in\R$. Then

$Tf(x) = x(a+bx+cx^2)+ (b+2cx) = b + (a+2c)x + bx^2 + c x^3 \in\mathcal{P}_3(\R).$

If $Tf=0$ then the four coefficients of $Tf$ must vanish, so we get four equations for $a,b,c$:

$\bigl\{b=0, \quad a+2c=0, \quad b=0, \quad c=0\bigr\} \implies a=b=c=0.$

Therefore the only $f\in\mathcal{P}_2(\R)$ that satisfies $Tf=0$ is the zero polynomial: $N(T) = \{0\}$. It follows that $T$ is injective (one-to-one). There is no basis for the null space.

The Range. The rank+nullity theorem implies $\dim R(T) = \dim \mathcal{P}_2(\R)-\dim N(T) =3-0=3$. So $R(T)$ is a 3-dimensional subspace of the four dimensional space $\mathcal{P}_3(\R)$. This implies $T$ is not surjective (onto).

To find a basis for $R(T)$ choose a basis for $\mathcal{P}_2(\R)$, such as $\{1, x, x^2\}$. Since $T$ is injective, $\{T1, Tx, Tx^2\}$ is a basis for the range of $T$. We find $T1=x$, $Tx=x^2+1$, $Tx^2=x^3+2x$. Therefore $\{x, 1+x^2, 2x+x^3\}$ is a basis of $R(T)$.

9

a. $T(a_1, a_2)= (1, a_2)$ is not linear because $T(0,0)=(1,0)\neq (0,0)$.

b. $T(a_1, a_2)= (a_1, a_1^2)$ is not linear because if $a=(1,0)$ then $Ta=(1,1)$ and $T(2a)=(2, 4)\neq(2,0)= 2a$

e. $T(a_1, a_2)= (a_1+1, a_1)$ is not linear becausev$T(0,0)=(1,0)\neq (0,0)$.

10

Let $u=\binom{1}{0}$, $v=\binom{1}{1}$. Then $\{u,v\}$ is linearly independent, because for all $a,b\in\R$

$au+bv=0\implies \binom{a+b}{b}=0\implies a=b=0.$

Since $\R^2$ is 2-dimensional, $\{u, v\}$ a basis for $\R^2$. We can therefore write every vector $x\in\R^2$ as a linear combination of $u$ and $v$.

Compute $T\binom23$. We are given $Tu=\binom14$ and $Tv=\binom25$. To compute $T\binom23$, write $\binom23$ as a linear combination of $u$ and $v$:

$\binom23 = au+bv = \binom{a+b}{b} \implies \begin{aligned} a+b&=2 \\ b&=3 \end{aligned} \implies a=-1, b=3.$

Therefore $\binom23 = -u+3v$, and hence

$T\binom23 = T(-u+3v)=-Tu+3Tv = -\binom14+3\binom25 =\binom5{11}.$

Is $T$ one-to-one? To see if $T$ is injective (one-to-one) we compute $N(T)$, i.e. we find all solutions of $Tx=0$. If $Tx=0$ then there exist $a,b\in\R$ with $x=au+bv$, because $\{u,v\}$ is a basis of $\R^2$. Using linearity of $T$ again, we get

$Tx = T(au+bv) = aTu+bTv = a\binom14+b\binom25 = \binom{a+2b}{4a+5b}.$

It follows that

$Tx=0\implies \begin{aligned} a+2b&=0 \\ 4a+5b&=0 \end{aligned} \implies a=b=0 \implies x=0.$

This shows that $N(T)$ contains only the zero vector. By the injectivity theorem $T$ is one-to-one.

11

Let $u=\binom11$, $v=\binom23$. Then $\{u,v\}$ is a basis for $\R^2$.
We define $Tu= \left(\begin{smallmatrix}1 \\ 0 \\ 2 \end{smallmatrix}\right)$ and $Tv =\left(\begin{smallmatrix}1 \\ -1 \\ 4 \end{smallmatrix}\right)$.

For any vector $x\in\R^2$ there are $a,b\in\R$ such that $x=au+bv$. The coefficients $a,b$ are uniquely determined by the vector $x$. Define $Tx$ by setting

$Tx=T(au+bv)=aTu+bTv = a\left(\begin{smallmatrix} 1 \\ 0 \\ 2 \end{smallmatrix}\right)+ b\left(\begin{smallmatrix} 1 \\ -1 \\ 4 \end{smallmatrix}\right)$

To prove that this definition makes $T$ linear, let $x=au+bv$, and $y=cu+dv$.
Then

$\begin{aligned} x+y&=(au+bv)+(cu+dv)= (a+c)u+(b+d)v \implies \\ T(x+y)&=T\left((a+c)u+(b+d)v\right)= (a+c)Tu + (b+d)Tv \end{aligned}$

On the other hand

$\begin{aligned} Tx=aTu+bTv, \qquad &Ty=cTu+dTv \\ \implies Tx+Ty &= aTu+bTv+cTu+dTv \\ &= (a+c)Tu + (b+d)Tv. \end{aligned}$

Therefore $T(x+y)=Tx+Ty$. A similar computations shows that $T(ax)=aTx$ for all $a\in\R$ and all $x\in \R^2$.

To compute $T\binom8{11}$ find $a, b\in\R$ such that $\binom{8}{11} = au+bv$. This leads to the following equations:

$\binom{8}{11} = au+bv = \binom{a+2b}{a+3b} \implies \begin{aligned} a+2b&=8 \\ a+3b&=11 \end{aligned} \implies a=2, b=3.$

Then

$T\binom{8}{11} = 2Tu+3Tv =2\left(\begin{smallmatrix} 1 \\ 0 \\ 2 \end{smallmatrix}\right)+ 3\left(\begin{smallmatrix} 1 \\ -1 \\ 4 \end{smallmatrix}\right) =\left(\begin{smallmatrix} 5 \\ -3 \\ 16 \end{smallmatrix}\right) .$

13 (write a careful proof)

To prove: Let $V$ and $W$ be vector spaces, let $T: V → W$ be linear, and let $\{w_1, w_2, \dots ,w_k\}$ be a linearly independent subset of $R(T)$. Prove that if $S = \{v_1, v_2,\dots ,v_k\}$ is chosen so that $T(v_i) = w_i$ for $i = 1, 2,\dots ,k$, then $S$ is linearly independent.

Proof. We can reformulate this as follows: if $T:V\to W$ is linear, and if for certain vectors $\{v_1,\dots, v_k\}\subset V$ the vectors $\{Tv_1, \dots, Tv_k\}$ are linearly independent, then $\{v_1, \dots, v_k\}$ also is linearly independent.

Suppose $c_1v_1+\cdots+c_kv_k =0$. Then, by linearity of $T$, we have

$0=T\bigl(c_1v_1+\cdots+c_kv_k\bigr) =c_1Tv_1 + \cdots + c_k Tv_k$

Since we know that $\{Tv_1, \dots, Tv_k\}$ is independent it follows that $c_1=c_2=\cdots=c_k=0$. Therefore $\{v_1, \dots, v_k\}$ is independent.

20 (write a careful proof)

To prove: Let $V$ and $W$ be vector spaces with subspaces $V_1$ and $W_1$, respectively. If $T: V → W$ is linear, prove that

1. $T(V_1)$ is a subspace of $W$, and
2. $\{x ∈ V \mid T(x) ∈ W_1\}$ is a subspace of $V$.

Proof of 1. By definition $T(V_1) = \{Tv \mid v\in V_1\}$. We verify that this set contains the zero vector, is closed under addition and scalar multiplication:

• Since $T(0)=0$, we have $0\in T(V_1)$.
• If $x, y\in T(V_1)$ then there are $u, v\in V_1$ such that $x=Tu$ and $y=Tv$. Hence $x+y= Tu+Tv=T(u+v)$. By definition $T(u+v)\in T(V_1)$, so $x+y\in T(V_1)$.
• If $x\in T(V_1)$ and $a\in\mathbb F$, then there is a $u\in V_1$ with $x=Tu$. It follows that $ax=aT(u)=T(au)\in T(V_1)$.

Thus $T(V_1)$ is a subspace.

Proof of 2. Let $L=\{x ∈ V\mid T(x) ∈ W_1\}$. To show that $L$ is a subspace, we again verify that $0\in L$, and that $L$ is closed under addition and scalar multiplication.

• $T$ is linear so $T(0_V)=0_W$. $W_1\subset W$ is a subspace, so $0_W\in W_1$. Hence $T(0_V)\in W_1$, and, by definition, $0_V\in L$.
• If $x, y\in L$ then $Tx, Ty\in W_1$. $W_1\subset W$ is a subspace, so $Tx+Ty\in W_1$. Since $T$ is linear we have $T(x+y)=Tx+Ty$. Therefore $T(x+y)\in W_1$, and thus $x+y\in L$.
• If $a\in\mathbb F$ and $x\in L$ then $T(x)\in W_1$.
  $W_1\subset W$ is a subspace, so $a\,T(x)\in W_1$.
  Since $T$ is linear we have $aT(x)=T(ax)$.
  Therefore $T(ax)\in W_1$, and thus $ax\in L$.

It follows that $L\subset W$ is a subspace.

problem 26 from section 1.6, using theorems about linear transformations.

Consider the function $T:\mathcal P_n(\R)→\R$ defined by $T(f)=f(a)$.

Show that T is linear. If $f, g\in\mathcal{P}_n(\R)$ then $T(f+g)=(f+g)(a)$, i.e. $T(f+g)$ is the value of the function $f+g$ at $a$. By definition of $f+g$ this is $f(a)+g(a)$. Thus we have $T(f+g)=(f+g)(a)=f(a)+g(a)=Tf+Tg$.

If $t\in\R$ and $f\in \mathcal{P}_n(\R)$ then $T(tf) = (tf)(a)= t(f(a)) = t\, Tf$.

Show that T is onto. For every $c\in \R$ consider the constant polynomial $f(x)=c$. Then $Tf=f(a)=c$, so $c\in R(T)$. Hence $R(T)=\R$; $T$ is onto.

Show that the set $L=\{f∈\mathcal P_n(\R)\mid f(a)=0\}$ is a linear subspace of $\mathcal P_n(\R)$ and use the rank+nullity theorem to compute its dimension. $L$ is the null space of $T$. Therefore $L$ is a linear subspace of $\mathcal{P}_n(\R)$. The rank+nullity theorem implies

$\dim L + \dim R(T) = \dim \mathcal{P}_n(\R).$

Since $\dim \mathcal{P}_n(\R) = n+1$ and since $R(T)=\R$ is one dimensional, we conclude $\dim L = n+1-1=n$.