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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?

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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.

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    $\begingroup$ 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. $\endgroup$
    – J.G.
    Sep 21 '19 at 17:53

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