$tr(A^*A)=tr(A^2)$ $ \Rightarrow$ $A$ is Hermitian matrix Let $A \in {M_n}(\mathbb{C})$  and assume $\mathrm{tr}(A^*A)=\mathrm{tr}(A^2)$. Why is $A$ a Hermitian matrix?
 A: Here is another (simple) proof that uses solely the properties of the trace operator. First, we recall that $tr(B^*B)=0$ if and only if $B=0$.
Second, the assumption $tr(A^*A)=tr(A^2)$ implies $tr(AA^*)=tr(A^2)$,
and by linearity 
$$tr(A(A^*-A))=0.$$
Moreover, $tr(B^*) = \overline{tr(B)}$, hence
$$
tr(A^*(A-A^*))=0.
$$
Then we compute
$$
tr((A-A^*)(A-A^*)^*)=tr((A-A^*)(A^*-A)) = tr(A(A^*-A)) - tr(A^*(A^*-A))=0.
$$
Hence $A=A^*$.
A: Do you remember the following?
If $M$ is a complex $n\times n$ matrix, then 
$$tr(M^*M)=0 \Rightarrow M=0.$$
And, perhaps independently, can you think of some decomposition of a general square complex matrix into components that may be of use in this problem?
Also my advice is: keep at hand all well-known properties of the trace.
A: The bilinear form $\langle A,B\rangle:=\text{tr}(A^*B)$ defines an inner product as you may check, and by the Cauchy-Schwarz inequality $\displaystyle|\langle B,A\rangle|=\|B\|\|A\|$ if and only if $A=\lambda B$ for some scalar $\lambda\,.$ Hence taking $B=A^*$ we have that $\text{tr}(A^2)=\text{tr}(A^*A)$ if and only if $A=\lambda A^*\implies \text{tr}(A^*A)=\text{tr}(A^2)=\lambda \text{tr}(A^*A)\,,$ so that $\lambda =1$ (assuming $A\not =0$). Hence $A=A^*\,.$
A: Apply Schur decomposition to $A = U^*TU$, where $U$ is a unitary matrix, $T$ an upper triangular matrix where the diagonal entries are the eigenvalues of $A$. 
$${\rm tr}(A^*A)={\rm tr} (T^*T)=\sum\limits_{\lambda} |\lambda|^2+\sum\limits_{j>i}|T_{ij}|^2$$ 
while 
$${\rm tr}(A^2)={\rm tr}(T^2)=\sum\limits_{\lambda} \lambda^2,$$
where $\lambda$ runs through all the eigenvalues of $A$. Obviously $$\sum\limits_{\lambda} |\lambda|^2+\sum\limits_{j>i}|T_{ij}|^2\ge \Big|\sum\limits_{\lambda} \lambda^2\Big|.\tag 1$$ 
By the triangular inequality $$\sum\limits_{\lambda} |\lambda|^2\ge \Big|\sum\limits_{\lambda} \lambda^2\Big|, \tag 2$$ 
the equality in Eq. (1) holds only if $T_{ij}=0,\,\forall i<j$. Moreover, for the equality in Eq. (2) to hold, all summands $\lambda^2$'s are necessarily colinear. It is only possible when $\lambda^2$ are positive considering the positivity of the left hand side of the equation. Therefore, all $\lambda$'s are real and $T$ is diagonal. In other words, $A$ is Hermitian.
