Zero determinant

Let $A \in \mathcal{M}_n \left( \mathbb{C} \right)$ such that

$A \det A + A^* \det A^*=i \left( A+ A^* \right)$

Prove that $\det A=0$ if $n$ is odd.

Solution

Let $\det A = z$ . Then

$zA + \bar{z}A^{\ast} = i(A+A^{\ast})$

Taking conjugate transpose we also have that

$\bar{z}A^{\ast} + zA = -i(A^{\ast} + A)$

Hence $A + A^{\ast} = 0$ . However it also holds $zA + \bar{z}A^{\ast} = 0$ . Combiming these two we get that

$(z-\bar{z})A = 0$

If $A = 0$ we are done. Otherwise $z$ is real. In that case we have

$z = \det A = \det(-A^{\ast}) = (-1)^n\det A^\ast = -\bar{z} = -z$

since $n$ is odd. Hence $\det{A} = z = 0$ as wanted.

Does there exist an enumeration?