Let $f : M_n → C$ be a linear map. Show that there exists a matrix $B$ ∈ $M_n$ such that $f(A) = tr(BA)$, $∀ A ∈ M_n$ Let $M_n$ be the vector space of $n×n$ complex matrices. Let $f : M_n → C$ be a linear
map. Show that there exists a matrix $B$ ∈ $M_n$ such that
$f(A) = tr(BA)$, $∀ A ∈ M_n$,
where tr is the normalized trace $(tr(1) = 1)$ on $M_n$.
Please give me an intuition how to solve it. I have no idea how to convert a linear map into form of trace of some matrix.
Thank you
 A: The space of linear functionals on $M_n$ is generated by the "entry-functionals"
$$
f_{ij}:M_n\to\mathbb C: A\mapsto A_{ij}
$$
where for a given $A\in M_n$ we denote $A_{ij}$ its $(i,j)$-entry.
Hint: what is $tr(BA)$ when $B=E_{ij}$, namely $B$ has only $0$ entries, except for the entry $i,j$ which is $1$?
A: Intuition
$$\langle A, B \rangle = \operatorname{tr}(\overline{A}^T B)$$ is an inner product over the linear space of square complex matrices of dimension $n$, named Frobenius inner product.
Then apply Riesz representation theorem. This is the natural way if you know Frobenius inner product.
If not, you can prove that any coordinate linear form over the linear space of square matrices of dimension $n$ can be represented by $M \mapsto \operatorname{tr}(A M)$. And then remember that those coordinate linear forms are a basis of the dual space. This is true whatever the base field is.
A: This is equivalent to saying $End(V)\otimes End(V)\xrightarrow{\circ}End(V)\xrightarrow{tr}  k$ is a non-degenerate pairing. Recall that $tr\colon End(V)\to k$ is the composition of $End(V)\xleftarrow\sim V^\vee\otimes V\xrightarrow{ev}k $, where the left map sends $v^\vee\otimes v\mapsto v^\vee(-)v$ and the right map is $v^\vee\otimes v\mapsto v^\vee(v)$.
There is a commutative diagram:
$\require{AMScd}$
\begin{CD}
V^\vee\otimes V\otimes V^\vee\otimes V @>{1\otimes ev\otimes 1}>> V^\vee\otimes V @>{ev}>> k,\\
@V{\sim}VV @V{\sim}VV & @V{=}VV\\
End(V)\otimes End(V)@>{-\circ-}>> End(V) @>{tr}>>k,
\end{CD}
where the top composition is $v^\vee\otimes v\otimes w^\vee\otimes w\mapsto v^\vee(w)w^\vee(v)$. Thus the induced map on the dual by the bilinear form on $V^\vee\otimes V$ is
$$\begin{align*}
V^\vee\otimes V&\to (V^\vee\otimes V)^\vee\xrightarrow{\sim} V^{\vee\vee}\otimes  V^{\vee}\cong V\otimes V^\vee\\
v^\vee\otimes v&\mapsto (w^\vee\otimes w\mapsto w^\vee(v)v^\vee(w))\mapsto v\otimes v^\vee,
\end{align*}
$$
which just swaps the factors, so is clearly an isomorphism.
