Suppose $V$ is the vector space consisting of all $n \times n$ square matrices. How can we find the form of a linear functional on $V$ ? Hofmann's book says (in exercise 8 of section 7 of chapter 3) that every functional on $V$ has the form ${\rm tr}(B^t A)$ for some matrix $B$. Why? and how?
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Let $E_{i,j}$ denote the matrix whose $(i,j)^{th}$ entry is 1, and all others are zero. Let $f_{i,j}$ denote the dual basis; then it suffices to prove this result for these $f_{i,j}$'s.
Consider $A = \sum a_{i,j} E_{i,j} \in V$, then $$ f_{k,l}(A) = a_{k,l} = tr(E_{k,l}^t A) $$
Prahlad Vaidyanathan
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The vector space of square matrices of order $n$ is isomorphic to $\mathbb R^{n^2}$. This is directly followed by $\text{vec}$ operation which sends a matrix $A$ to a vector $\text{vec}(A)$. According to Riesz representation theorem, there exists a unique vector $v=\text{vec}(B)\in\mathbb R^{n^2}$ such that for any $u=\text{vec}A$, we have $$\left<u,v\right>=\sum_{k=1}^{n^2}u_kv_k=\sum_{i=1}^n\sum_{j=1}^nA_{ij}B_{ji}=\text{tr}B^TA$$
Shuchang
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There's a more general result. Given a $\Bbb K-$vector space ($\Bbb K=\Bbb R$ or $\Bbb C$) $E$ and an inner product $\langle\cdot \mid \cdot \rangle:E\times E \to \Bbb K$, the homomorphism $$\varphi:\begin{matrix}E \to E^*=\mathcal F\left(E,\Bbb K\right)\\ x\mapsto \left(y\mapsto \langle x\mid y\rangle\right)\end{matrix}$$
is injective. In the case where $E$ is finite dimensional, $\dim E^*=\dim \mathcal F\left(E,\Bbb K\right)=\dim E \times \dim \Bbb K=\dim E \times 1 = \dim E$ so since $\varphi $ is injective, it is bijective.
Now in your case, $E=\mathcal M_n\left(\Bbb K\right)$ and $\langle x\mid y\rangle = \operatorname{tr}\left(^tB A\right)$.
xavierm02
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