# Sum of two hermitian matrices

How can I prove that any square matrix $$M_n(\mathbb{C})$$ can be written in the form of $$A = B + iC$$ with both B and C being hermitian matrices?

In exactly the same way you do it for expressing complex numbers as $$a=b+ic$$ for real numbers $$b,c$$: $$A = \frac{A + A^{\dagger}}{2} + \frac{A-A^{\dagger}}{2} = \frac{A + A^{\dagger}}{2} + i\frac{A-A^{\dagger}}{2i} .$$ It remains to check that $$\frac{1}{2}(A+A^{\dagger})$$ and $$\frac{1}{2i} (A-A^{\dagger})$$ are Hermitian: take the Hermitian conjugate and see what happens.
• And even if you didn't think of that decomposition, you can note if $A=B+iC$ with $B,\,C$ Hermitian then $A^\dagger=B-iC$, and now we just need to solve simultaneous equations. In fact, this implies the decomposition is unique.