Suppose we have a matrix $A \in M_{n\times n}(\mathbb C)$. It is well-known that we can write $A$ as $$ A=H(A)+iK(A) $$ where $H(A)=\frac 1 2 (A+\bar A^t)$ is an Hermitian matrix and $K(A)= \frac 1 2 (A-\bar A^t)$ is skew Hermitian.
I want to proceed in the opposite direction: given an Hermitian matrix $B$, can I always find a matrix $A$ such that $B=H(A)$? I am tempted to say yes, but I can't find a convincing argument.
Yes this is true, just take $A = B$. Since $B$ is Hermitian we have $\overline{B}^t = B$ and hence $$H(B) = \frac 1 2 (B + \overline{B}^t) = \frac 1 2 (B+B) = B$$