Determinant of a special block matrix in terms of a singular matrix

Question

I have a matrix $A$ with $\det A = 0$. How can one prove that for $Z = \begin{pmatrix} \Re[A] & -\Im[A] \\ \Im[A] & \Re[A] \end{pmatrix}$, is such that $\det Z =0$?

Answer 1

Since $\det A = 0$ i, we know that there is some non zero $x$ such that $Ax = 0$.

Then, $y = \begin{pmatrix} \Re[x] \\ \Im [x]\end{pmatrix} $ is non zero and : \begin{align} Zy &= \begin{pmatrix} \Re [A] & -\Im[A]\\ \Im[A] & \Re[A] \end{pmatrix} \begin{pmatrix} \Re[x] \\ \Im [x]\end{pmatrix} \\ &= \begin{pmatrix} \Re[A]\Re[x] - \Im[A]\Im[x] \\ \Re[A]\Im[x] + \Im[A]\Re[x] \end{pmatrix}\\ &= \begin{pmatrix} \Re[Ax]\\ \Im[Ax] \end{pmatrix} \\ &= 0 \end{align}

The next-to-last step follows from : \begin{align} Ax &= (\Re[A] + i\Im[A])(\Re[x] + i\Im[x]) \\&= (\Re[A]\Re[x] - \Im[A]\Im [x] ) + i (\Re[A]\Im[x] + \Im[A]\Re[x]) \end{align} Therefore, $\det Z = 0$