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## Solution to Linear Algebra Hoffman & Kunze Chapter 3.7

#### Exercise 3.7.1

Let $F$ be a field and let $f$ be the linear functional on $F^2$ defined by $f(x_1,x_2)=ax_1+bx_2$. For each of the following linear operators $T$, let $g=f^tf$, and find $g(x_1,x_2)$.

(a) $T(x_1,x_2)=(x_1,0)$;
(b) $T(x_1,x_2)=(-x_2,x_1)$;
(c) $T(x_1,x_2)=(x_1-x_2,x_1+x_2)$.

Solution:

(a) $g(x_1,x_2)=T^tf(x_1,x_2)=f(T(x_1,x_2))=f(x_1,0)=ax_1$.

(b) $g(x_1,x_2)=T^tf(x_1,x_2)=f(T(x_1,x_2))=f(-x_2,x_1)=-ax_s+bx_1$.

(c) We have \begin{align*}g(x_1,x_2)&=T^tf(x_1,x_2)=f(T(x_1,x_2))=f(x_1-x_2,x_1+x_2)\\&=a(x_1-x_2)+b(x_1+x_2)=(a+b)x_1+(b-a)x_2.\end{align*}

#### Exercise 3.7.2

Let $V$ be the vector space of all polynomial functions over the field of real numbers. Let $a$ and $b$ be fixed real numbers and let $f$ be the linear functional on $V$ defined by
$$f(p)=\int_a^bp(x)dx.$$ If $D$ is the differentiation operator on $V$, what is $D^tf$?

Solution: Let $p(x)=c_0+c_1x+\cdots+c_nx^n$. Then
\begin{alignat*}{1}
D^tf(p)=&f(D(p))\\
=&f(c_1+2c_2x+3c_3x^2+\cdots+nc_nx^{n-1})\\
=&c_1+c_2x^2+\cdots c_nx^n\mid_0^b\\
=&p(b)-p(a).
\end{alignat*}

#### Exercise 3.7.3

Let $V$ be the space of all $n\times n$ matrices over a field $F$ and let $B$ be a fixed $n\times n$ matrix. If $T$ is the linear operator on $V$ defined by $T(A)=AB-BA$, and if $f$ is the trace function, what is $T^tf$?

Solution: By exercise 3 in section 3.5, we know trace$(AB)$ $=$ trace$(BA)$. Thus \begin{align*}T^tf(A)&=f(T(A))=\text{trace}(AB-BA)\\&=\text{trace}(AB)-\text{trace}(BA)=0.\end{align*}

#### Exercise 3.7.4

Let $V$ be a finite-dimensional vector space over the field $F$ and let $T$ be a linear operator on $V$. Let $c$ be a scalar and suppose there is a non-zero vector $\alpha$ in $V$ such that $T\alpha=c\alpha$. Prove that there is a non-zero linear functional $f$ on $V$ such that $T^tf=cf$.

Solution: Consider the operator $U=T-cI$. Then $U(\alpha)=0$ so $\text{rank}(U)<n$. Therefore $\text{rank}(U^t)<n$ as an operator on $V^*$. It follows that there’s a $f\in V^*$ such that $U^t(f)=0$. Now $U^t=T^t-cI$, thus $T^t(f)=cf$.

#### Exercise 3.7.5

Let $A$ be an $m\times n$ matrix with real entries. Prove that $A=0$ if and only if trace$(A^tA)=0$.

Solution: Suppose $A$ is the $m\times n$ matrix with entries $a_{ij}$ and $B$ is the $n\times k$ matrix with entries $b_{ij}$. Then the $i,j$ entry of $AB$ is
$$\sum_{k=1}^na_{ik}b_{kj}.$$Substituting $A^t$ for $A$ and $A$ for $B$ we get the $i,j$ entry of $A^tA$ is (note that $A^tA$ has dimension $n\times n$)
$$\sum_{k=1}^ma_{ki}a_{kj}.$$Thus the diagonal entries of $A^tA$ are
$$\sum_{k=1}^ma_{ki}a_{ki}$$for $i=1,\dots,n$. Thus the trace is
$$\sum_{i=1}^n\sum_{k=1}^ma_{ki}^2.$$If all $a_{ij}\in\mathbb R$ then this sum is zero if and only if each $a_{ij}=0$ because every one of them appears in this sum.

#### Exercise 3.7.6

Let $n$ be a positive integer and let $V$ be the space of all polynomial functions over the field of real numbers which have degree at most $n$, i.e., functions of the form
$$f(x)=c_0+c_1x+\cdots+c_nx^n.$$Let $D$ be the differentiation operator on $V$. Find a basis for the null space of the transpose operator $D^t$.

Solution: The null space of $D^t$ consists of all linear functionals $g:V\rightarrow \mathbb R$ such that $D^tg=0$, or equivalently $g\circ D(f)=0$ $\forall$ $f\in V$. Now $g(a_0+a_1x+\cdots+a_nx^n)=c_0a_0+c_1a_1+\cdots+c_na_n$ for some consants $c_0,\dots,c_n\in\mathbb R$. So \begin{align*}g\circ D(a_0+a_1x+\cdots+a_nx^n)&=g(a_1+2a_2x+\cdots+na_nx^{n-1})\\&=\sum_{i=0}^{n-1}(i+1)c_ia_{i+1}.\end{align*}

This sum $\sum_{i=0}^{n-1}(i+1)c_ia_{i+1}$ equals zero for all vectors $$(a_0,a_1,\dots,a_n)\in\mathbb R^n$$ if and only if $c_0=c_1=\cdots=c_{n-1}=0$. While $c_n$ can be anything. Therefore the null space has dimension one and a basis is given by taking $c_n=1$, which gives the function $g:\sum a_ix^i\mapsto a_n$, the projection onto the $x^n$ coordinate.

#### Exercise 3.7.7

Let $V$ be a finite-dimensional vector space over the field $F$. Show that $T\rightarrow T^t$ is an isomorphism of $L(V,V)$ onto $L(V^*,V^*)$.

Solution: Choose a basis $\mathcal B$ for $V$. This gives an isomorphism $L(V,V)\rightarrow M_n$ the space of $n\times n$ matrices. And the dual basis $\mathcal B’$ gives an isomorphism $L(V^*,V^*)\rightarrow M_n$. Now we know by Theorem 23 that the following diagram of functions commutes. This means if we start at $L(V,V)$ and follow two functinos to the $M_n$ in the bottom left, it doesn’t matter which way around the diagram we go, we end up at the same place.
\begin{alignat*}{3}
&L(V,V) &\longrightarrow & M_n\\
\text{transpose}&\downarrow & &\downarrow \text{transpose}\\
&L(V^*,V^*) &\longrightarrow & M_n
\end{alignat*}Both horizontal arrows are isomorphisms (by Theorem 12, page 88). Now clearly transpose on matrices is a one-to-one and onto function from the set of matrices to itself. Also $(rA)^t=rA^t$ and $(A+B)^t=A^t+B^t$ for any two $n\times n$ matrices $A$ and $B$. Thus transpose is also a linear transformation. Thus transpose is an isomorphism. Therefore three of the arrows in this diagram are isomorphisms and it follows that the fourth arrow must also be an isomorphism.

#### Exercise 3.7.8

Let $V$ be the vector space of $n\times n$ matrices over the field $F$.

(a) If $B$ is a fixed $n\times n$ matrix, define a function $f_B$ on $V$ by $F_B(A)=\text{trace}(B^tA)$. Show that $F_B$ is a linear functional on $V$.
(b) Show that every linear functional on $V$ is of the above form, i.e., is $f_B$ for some $B$.
(c) Show that $B\rightarrow f_B$ is an isomorphism of $V$ onto $V^*$.

Solution:

(a) This follows from the fact that the trace function is a linear functional and left multiplication by a matrix is a linear transformation from $V$ to $V$. In other words \begin{align*}F_B(cA_1+A_2)&=\text{trace}(B^t(cA_1+A_2))\\&=\text{trace}(cB^tA_1+B^tA_2)\\&=c\cdot\text{trace}(B^tA_1)+\text{trace}(B^tA_2)\\&=c\cdot F_B(A_1)+F_B(A_2).\end{align*}

(b) Let $f:V\rightarrow F$ be a linear functional. Let $A=(a_{ij})\in V$. Then

f(A)=\sum_{i,j=1}^nc_{ij}a_{ij}
\label{qef2f3}
for some fixed constants $c_{ij}\in F$.

Now let $B=(b_{ij})\in V$ be any matrix. Then the $i,j$ element of $B^tA$ is $\sum_{k=1}^nb_{ki}a_{kj}$. Thus

\text{trace}(B^tA)=\sum_{i=1}^n\sum_{k=1}^nb_{ki}a_{ki}.
\label{qef2f3fdw}
Comparing (\ref{qef2f3}) and (\ref{qef2f3fdw}) we see each $a_{ij}$ apperas exactly once in each sum. So setting $b_{ki}=c_{ki}$ for all $i,k=1,\dots,n$ we get the appropriate matrix $B$ such that $\text{trace}(B^tA)=f$.

(c) Let $F$ be the function $F:V\rightarrow V^*$ such that $F(B)=f_B$. Part (a) shows this function is into $V^*$. Part (b) shows it is onto $V^*$. We must show it is linear and one-to-one. Let $r\in F$ and $B_1,B_2\in V$. Then $$(rB_1+B_2)^tA=(rB_1^t+B_2^t)A=rB_1^tA+B_2^tA.$$ We know the trace function itself is linear. Thus $F$ is linear. Now suppose trace$(B^tA)=0$ $\forall$ $A\in V$. Fix $i,j\in\{1,2,\cdots,n\}$. Let $A$ be the matrix with a one in the $i,j$ position and zeros elsewhere. Then by the proof of part (b) we know that trace$(B^tA)=b_{ij}$. So if $F(A)=0$ then $b_{ij}=0$. Thus $B$ must be the zero matrix. Thus $F$ is one-to-one. It follows that $F$ is an isomorphism.