Let $M$ be an $n \times n$ complex matrix. Prove that there exist Hermitian matrices $A$ and $B$ such that $M = A + iB$ The back of the book provides the following explanation:
If $M = A + iB$, then $\overline{M^t} = \overline{A^t} − i\overline{B^t} = A − iB$. 
So we should take
$A = \frac{1}{2}(M + \overline{M^t})$ and $B = \frac{1}{2i}(M − \overline{M^t})$, which are of course both Hermitian.
I thought that $\overline{M^t} = \overline{A^t} − i\overline{B^t} = A − iB$ only if $A$ and $B$ are symmetric? So that $A^t$ = $A$ and $B^t = B$.
Then for the second part of the explanation, what makes $A = \frac{1}{2}(M + \overline{M^t})$ Hermitian? 
 A: The derivation of the formulas for $A$ and $B$ works after we assume that such hermitian matrices indeed exist. So, it should read like

Suppose that there are hermitian $A$ and $B$ such that $M=A+iB$. Then ...

Now, once we got the formulas, we can prove that they indeed give us hermitian matrices. If
$$A = \frac{1}{2}(M+M^*)$$
then
$$A^* = \frac{1}{2}(M^* + M^{**}) = \frac{1}{2}(M^* + M) = A$$
And similarily for $B$ (note how the signs cancel out):
$$B^* = \frac{1}{-2i}(M^* - M^{**}) = \frac{1}{-2i}(M^* - M) = B$$
NB: I use $A^*$ for $\overline{A^t}$

If you are interested in some more high-level view of the problem, observe that the hermitian transpose operator is self-inverse, and so provides a representation of the group $C_2$ in the space of matrices. General representation theory of $C_2$ tells us that the whole space decomposes into a sum of two subspaces:


*

*Matrices such that $A^*=A$

*Matrices such that $A^*=-A$


The first subspace is the space of hermitian matrices, the second is the space of anti-hermitian matrices.
The only thing that remains is the fact that $A$ is hermitian $\Leftrightarrow$ $iA$ is anti-hermitian.
