Question envolving trace and linear functionals. I need to prove that $F^{n\times n} \approx (F^{n\times n})^*$ (dual space of $F^{n\times n}$) by isomorphism.
My Attempt:
Lets $f_A:F^{n\times n} \to F$, $f_A(X)= Tr(AX)$, it is obvious that $f \in (F^{n\times n})^*, \forall A \in F^{n \times n}.$ 
Now, define $T:F^{n\times n} \to (F^{n\times n})^* $ given by $T(A) = f_A.$ It's obvious that $T$ is linear. So I'm trying to prove that $KerT = \{0\}$, then $\dim F^{n\times n} = \dim (F^{n\times n})^* $ implies the result.
But if $A \in KerT$ then $T(A) = 0_f$, but $T(A) = f_A.$ So we have $f_A(X)= Tr(AX)= 0, \forall X \in F^{n \times n}$. How can I prove that the only matrix $A$ such that $Tr(AX) = 0$ for all X is $A=0??$
I can see the solution for $n=2$, that is,  $A=\begin{pmatrix}
a&b\\
c&d
\end{pmatrix} $, if we take $X_1 = \begin{pmatrix}
1&0\\
0&0
\end{pmatrix}, X_2= \begin{pmatrix}
0&1\\
0&0
\end{pmatrix}, X_3 = \begin{pmatrix}
0&0\\
1&0
\end{pmatrix}$ and $X_4= \begin{pmatrix}
0&0\\
0&1
\end{pmatrix}$.
Since $Tr(AX) =0, \forall X \in F^{2 \times 2}$, then
 $0=Tr(A X_i)$ implies $a=b=c=d=0$, so $A = 0.$
I imagine that in $F^{n \times n}$ we need to do the same thing, that is, $X_i$ is in the standard basis of $F^{n \times n}$. But i'm wondering, there is easier way??
 A: Your idea is fine. Checking $\text{Tr}(AX)=0$ for $X$ being each element of the standard basis of $F^{n \times n}$ will end up showing that each entry of $A$ is zero.

It may help to notice that the matrix structure is superficial. You are then just trying to show $F^{n^2} \cong (F^{n^2})^*$. Note that $\text{Tr}(A^\top X)$ is equivalent to the Euclidean dot product if you treat $A$ and $X$ as vectors of dimension $n^2$.
A: If $(V, \langle \cdot,\cdot\rangle)$ is a finite-dimensional vector space equipped with a non-degenerate pairing $\langle \cdot,\cdot\rangle \colon V\times V \to F$ (frequent case being where $F = \Bbb R$ and $\langle \cdot,\cdot\rangle$ is an inner product), then the map $$V \ni v \mapsto \langle v, \cdot \rangle \in V^*$$is an isomorphism. It is linear and indeed takes values in $V^*$ because $\langle \cdot,\cdot\rangle$ is bilinear. It is injective because $\langle \cdot,\cdot\rangle$ is non-degenerate. Then surjectivity follows since the dimension of $V$ is finite and we use that $\dim V =\dim V^*$.
Your situation is the particular case where $V=F^{n\times n}$ and $\langle A, B\rangle = {\rm tr}(AB^\top)$. Of course, the problem here would boil down to checking that this $\langle \cdot,\cdot\rangle$ is actually non-degenerate. But if $E_{k\ell}$ is the matrix with $1$ in the position $(k,\ell)$ and zero elsewhere, we have that $${\rm tr}(AE_{k\ell}^\top) = \sum_{i,j=1}^na_{ij}(E_{k\ell})_{ij}  = \sum_{i,j=1}^n a_{ij}\delta_{ki}\delta_{j\ell} = a_{k\ell}, $$and the result follows from $k$ and $\ell$ being arbitrary.
