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This is a problem our algebra teacher left for us. After researching on related topic, I have found out that it is a direct corollary of Wedderburn's Theorem, which is reads as follows: Suppose A is a simple finite F-algebra, then A is isomorphic to a matrix algebra over division ring D, which is a F-algebra , i.e A $simeq $$M_n(D)$. Since the division ring over $mathbbC$ should only be $mathbbC$ itself, if A is a simple associative algebra over $mathbbC$ , by Wedderburn's theorem, we can derive A $simeq $$M_n(mathbbC )$.

However, my professor expects a more elementary proof without generalizing to general occasions, say, to prove Wedderburn's theorem.

But until now, I have no clue about how to handle the original problem directly. I also check some literature. Some gave me a hint by considering B(a,b) = tr($L_aL_b$), which was proposed by Molien in 1892. But I don't know where to go on.

Thank you for your help in advance.

I try to formulate the proof of Wedderburn's theorem.

Lemma1:
Suppose A is a simple F-algebra, $mathcalI,mathcalJ$ are two minimal left ideal of A, then there exists $alpha in A$, s.t $mathcalI= mathcalJalpha$, thus $mathcalIcong mathcalJ$ holds.

Proof:
Noted that $mathcalIA, mathcalJA$ are two-sided ideal of A. Since A has no proper two-sided ideal, if $mathcalI,mathcalJ$ are not trivial, then $mathcalIA = mathcalJA =A$, so we can deduce that $mathcalIJA = mathcalIA=A$ and $mathcalIJneq 0$. Let $alpha in J$, then $mathcalIalpha subset mathcalJ$. Since $mathcalJ$ is a minimal left ideal and $ mathcalIalphaneq 0$, then $mathcalIalpha = mathcalJ$, then $mathcalIcong mathcalJ$.

Lemma2:
Suppose A is F-algebra, then $End_A(A) cong A^op$.

Proof: By defining a right multiplication action.

Lemma3:(suggested by @rschwieb)
Suppose $M_i,N_i,: i = 1,2,...,n$ are left R-module, then
$$
Hom_R( oplus_i=1^n M_i, N )cong oplus_i=1^n Hom_R(M_i, N)
$$
$$
Hom_R(M, oplus_i=1^n N_i )cong oplus_i=1^n Hom_R(M, N_i)
$$

Proof: By induction. When n=2, it is obvious.

Back to the proof of Wedderburn's theorem:
Since A is a simple algebra, we can write $A = oplus_i=1^n A_i$, where $A_i$ are the simple A-module. In other words, $A_i$ are the minimal left ideals of A. By lemma 1, the minimal left ideals of simple algebra A are isomorphic to each other. Thus, $A = oplus_i=1^n A_i cong oplus_i=1^n A_1$.
By lemma 2, we have $A^opcong End_A(A)=End_A(oplus_i=1^n A_1)$, where $End_A(A_1)$ is well-defined since $A_1$ is the left ideal of A. With the help of lemma 3, $$
End_A(oplus_i=1^n A_1) = Hom_A(oplus_i=1^n A_1, oplus_i=1^n A_1)cong oplus_i=1^n oplus_i=1^n Hom_A(A_1,A_1)
$$
$$= oplus_i=1^n oplus_i=1^n End_A(A_1)
$$ By
installing every item into the entry of a matrix, we have
$$
oplus_i=1^n oplus_i=1^n End_A(A_1)cong M_n( End_A(A_1))
$$
By Schur's lemma, since A1 is a simple A-module, $End_A(A_1)$ is a division ring, which is denoted by D. Namely, we have $A^op cong M_n(D)$. Then, $A cong M_n(D)^opcong M_n(D) $, as we desired, where n is the number of minimal left ideals of A and $D= End_A(A_1)$.