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
\begin{equation}
f(A)=\sum_{i,j=1}^nc_{ij}a_{ij}
\label{qef2f3}
\end{equation}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
\begin{equation}
\text{trace}(B^tA)=\sum_{i=1}^n\sum_{k=1}^nb_{ki}a_{ki}.
\label{qef2f3fdw}
\end{equation}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.
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