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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

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    $\begingroup$ Apply Riesz representation theorem to the hermitian space $M_n(\Bbb C)$ with $\langle M,N\rangle=tr(M^*\,N).$ $\endgroup$ Nov 13, 2022 at 16:59
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    $\begingroup$ For real numbers, see this post. Apply Riesz then the same way for Hermitian spaces, as Anne said. $\endgroup$ Nov 13, 2022 at 16:59
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    $\begingroup$ This is just a guess as to where I'd start, but what would happen if your took $A$ to be a matrix that was all $0$s except for a single $1$ in one position? I think you might be able to figure out $B$ one entry at a time. $\endgroup$
    – JonathanZ
    Nov 13, 2022 at 17:00

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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 $1$?

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    $\begingroup$ This is the same as JonathanZ supports MonicaC's previous comment. Note that it is better than Dietrich's or mine, in that it generalizes to any field of scalars (e.g. finite fields). And Riesz was overkill. $\endgroup$ Nov 13, 2022 at 17:07
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    $\begingroup$ @AnneBauval- Yeah, Riesz felt like a pretty big hammer to use, especially if it wasn't already known to the OP. I'm happy to have my hint rendered as a full answer with someone else adding all the details, assuming people think the original question isn't a PSQ. $\endgroup$
    – JonathanZ
    Nov 13, 2022 at 17:18
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    $\begingroup$ (@AnneBauval - And reading it more closely, this is definitely better than my comment. By using the $E_{ij}$ for $B$, I think you can explicitly write out $B$ by taking a linear combination of $tr(E_{ij} A)$ terms. I was suggesting computing the $b_{ij}$ terms and then having to show that they work. I've up voted this answer, but thanks for the acknowledgement.) $\endgroup$
    – JonathanZ
    Nov 13, 2022 at 17:31
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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.

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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.

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