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 $B \to f_B$ is an isomorphism of $V$ onto $V^∗$.
My attempt: (a) Let $c\in F$ and $P,Q\in V$. Then $f_B(c\cdot P+Q)$ $=\text{tr}(B^t\cdot (c\cdot P+Q))$ $=\text{tr}(B^t\cdot (c\cdot P)+B^t\cdot Q)$ $= \text{tr}(c\cdot (B^t \cdot P))+B^t\cdot Q)$. Since trace is a linear map, we have $\text{tr}(c\cdot (B^t \cdot P))+B^t\cdot Q)$ $=c\cdot \text{tr}(B^t\cdot P)+\text{tr}(B^t\cdot Q)$ $=c\cdot f_B(P)+f_B(Q)$. Thus $f_B(c\cdot P+Q)$ $=c\cdot f_B(P)+f_B(Q)$. Hence $f_B$ is a linear functional on $V$.
(b) Let $T:V\to F$ be a linear functional on $V$. So $T(E^{ij})\in F$, $\forall i,j\in J_n$, where $\{E^{ij}|1\leq i,j\leq n\}$ is standard basis of $V=M_n(F)$. Suppose $B$ $=(T(E^{ij}))$ $=(b_{ij})$. We claim $T=f_B$. Let $i,j\in J_n$. Then $f_B(E^{ij})$ $=\text{tr}(B^t\cdot E^{ij})$. Let $C=B^t\cdot E^{ij}$. Then $c_{pq}$ $=\sum_{k=1}^n b_{kp}\cdot (E^{ij})_{kq}$, for some $(p,q)\in J_n\times J_n$. So $\text{tr}(B^t\cdot E^{ij})$ $=\text{tr}(C)$ $=\sum_{p=1}^n c_{pp}$ $=\sum_{p=1}^n(\sum_{k=1}^nb_{kp}\cdot (E^{ij})_{kp})$. If $p\neq j$, then $(E^{ij})_{kp}=0$, $\forall k\in J_n$. So $\sum_{p=1}^n(\sum_{k=1}^nb_{kp}\cdot (E^{ij})_{kp})$ $=\sum_{k=1}^n b_{kj}\cdot (E^{ij})_{kj}$. If $k\neq i$, then $(E^{ij})_{kj}=0$. So $\sum_{k=1}^n b_{kj}\cdot (E^{ij})_{kj}$ $=b_{ij}\cdot (E^{ij})_{ij}$ $=b_{ij}\cdot 1$ $=b_{ij}$ $=T(E^{ij})$. Thus $f_B(E^{ij})$ $=T(E^{ij})$, $\forall (i,j)\in J_n\times J_n$. By theorem 1 section 3.1, $f_B=T$. Hence $\exists B\in V$ such that $f_B=T$.
(c) Define $g: V\to V^*$ such that $g(B)=f_B$. Let $c\in F$ and $P,Q\in V$. Then $g(c\cdot P+Q)$ $=f_{c\cdot P+Q}$. Let $A\in V=M_n(F)$. Then $f_{c\cdot P+Q}(A)$ $=\text{tr}((c\cdot P+Q)^t\cdot A)$. By exercise 7 section 3.7, $(c\cdot P+Q)^t$ $=c\cdot P^t+Q^t$. So $\text{tr}((c\cdot P+Q)^t\cdot A)$ $=\text{tr}((c\cdot P^t+Q^t)\cdot A)$ $=\text{tr}(c \cdot (P^t\cdot A)+Q^t\cdot A)$. Since trace is a linear map, we have $\text{tr}(c \cdot (P^t\cdot A)+Q^t\cdot A)$ $=c\cdot \text{tr}(P^t\cdot A)+\text{tr}(Q^t\cdot A)$ $c\cdot f_P(A)+f_Q(A)$ $=c\cdot f_P+f_Q(A)$. So $f_{c\cdot P+Q}$ $=c\cdot f_P+f_Q$. Thus $g(c\cdot P+Q)$ $=c\cdot g(P)+g(Q)$. Hence $g:V\to V^*$ is a linear map. Approach(1): By exercise 8(b) section 3.7, $\forall T\in V^*$, $\exists B\in V$ such that $g(B)$ $=f_B$ $=T$. Thus $g$ is surjective. Since $\mathrm{dim}(V)$ $= \mathrm{dim}(V^*)$ $=n^2$, we have $g$ is injective, by theorem 9 section 3.2. Hence $g$ is bijective. Approach(2): We can explicitly show $g$ is injective. If $g(B)=g(C)$, for some $B,C\in V$. Then $f_B=f_C$. So $f_B(A)$ $=f_C(A)$ $=\text{tr}(B^t\cdot A)$ $=\text{tr}(C^t\cdot A)$, $\forall A\in V$. In particular, $\text{tr}(B^t\cdot E^{ij})$ $=\text{tr}(C^t\cdot E^{ij})$, $\forall i,j\in J_n$. By exercise 8(b) section 3.7, $\text{tr}(B^t\cdot E^{ij})$ $=\text{tr}(C^t\cdot E^{ij})$ $=b_{ij}$ $=c_{ij}$, $\forall i,j\in J_n$. Thus $B=C$. Hence $g$ is injective. Another way to show injective, $N_g=\{0_V\}$. Let $B\in N_g$. Then $g(B)=f_B=0_{V^*}$. So $f_B(A)$ $=\text{tr}(B^t\cdot A)=0_F$, $\forall A\in V$. In particular, $\text{tr}(B^t\cdot E^{ij})=0_F$, $\forall i,j\in J_n$. By exercise 8(b) section 3.7, $\text{tr}(B^t\cdot E^{ij})$ $=b_{ij}$ $=0_F$, $\forall i,j\in J_n$. Thus $B=(b_{ij})=0_V$. Hence $N_g=\{0_V\}$. Is my proof correct?