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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.

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