Consider the following nonsingular block matrix: $$ M=\begin{pmatrix} A & B \\ -B^* & A^* \end{pmatrix} $$ where $A$ and $B$ are square matrices with the same dimension, $A^*$ and $B^*$ denote their complex conjugates.

Generating some random matrices for $A$ and $B$, I find that the determinant of $M$ is always a positive real number. Can anyone prove this fact, please?

  • $\begingroup$ Did you try to prove it yourself? $\endgroup$ – Rodrigo de Azevedo Oct 29 '19 at 11:53
  • $\begingroup$ @RodrigodeAzevedo Yes, I've tried. In fact, the question is from a problem I encountered, where $M$ is a similar transformation satisfying $M^{*-1}\Omega M=\Omega$ with $\Omega=\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix}$ (a generalization of unitary symplectic matrix). $\endgroup$ – zrysky Oct 29 '19 at 13:28
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    $\begingroup$ This is Lemma 2.2 in arxiv.org/abs/1505.04240 . $\endgroup$ – daw Oct 29 '19 at 14:14
  • $\begingroup$ @daw Many thanks. $\endgroup$ – zrysky Oct 29 '19 at 14:49

From this property of a block matrix, and assuming that $A^*$ and $B^*$ commute:

$$\text{Det}(M) = \text{Det}(AA^*+BB^*)$$

Because both $AA^*$ and $BB^*$ are symmetric semi-positive, then $\text{Det}(M) \geq0$. Naturally, if you generate matrices at random, it is likely that they'll both be symmetric positive definite, and the determinant will be positive. But is you plug in two null matrices, the determinant will be zero. If $A$ and $B$ are non-null but have the same non-empty null space, then $M$ won't be zero, but still $\text{Det}(M)$ will be zero.

  • $\begingroup$ @Mefitico From "this property", we can only obtain $\det(M)=\det(AA^*+BA^{*-1}B^*A^*)$ if $A$ and $B$ do not commute. Besides, $M$ has been assumed to be nonsingular, so $\det(M)\neq0$. $\endgroup$ – zrysky Oct 29 '19 at 13:02
  • $\begingroup$ @zrysky : From the question, I'm assuming you are unsure (with the conditions given) if the property actually holds. In fact, I've added one hypothesis (commutability) to prove that the determinant might be zero. Did you actually want to prove that it cannot be negative? $\endgroup$ – Mefitico Oct 29 '19 at 13:04
  • $\begingroup$ Indeed, the question comes from a problem I encountered, where $M$ denotes a similar transformation satisfying $M^{*-1}\Omega M=\Omega$ with $\Omega=\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix}$ (a generalization of symplectic matrix), so it has nonzero determinant. $\endgroup$ – zrysky Oct 29 '19 at 13:14

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