Interesting facts about the matrix $ A^\ast = \left(\begin{smallmatrix} A & -aI \\ a I & A \end{smallmatrix}\right)$ where $A$ is negative definite

Question

I have a fairly general question about facts of the following matrix

$$ A^\ast = \begin{pmatrix} A & -aI \\ a I & A \end{pmatrix}$$

where $A \in \mathbb{R}^{n \times n}$ is negative definite, $I$ is the $n$-dimensional identity matrix and $a>0$.

For example, from the above assumptions it's obvious that $A^\ast$ is negative definite. Using the answer to this question, the determinant of $A^\ast$ is given by $\det(A^\ast)=\det(A^2+a^2 I)$.

I'm looking for more "interesting" facts, like is there an easy way to compute $A^{\ast-1}$ given $A^{-1}$, can we say something about the eigenvalues of $A^\ast$ given the eigenvalues of $A$ etc.

Answer 1

In terms of the Kronecker product, your matrix is $$ A^* = I \otimes A + a \pmatrix{0&-1\\1&0} \otimes I. $$ Using this, we can deduce that for every eigenvalue $\lambda$ of $A$, $\lambda \pm ai$ are eigenvalues of $A^*$.

Note that, via the vectorization operator, we see that $A^*$ is the matrix associated with the operator $\Phi:\Bbb R^{n \times 2} \to \Bbb R^{n \times 2}$ defined by $$ \Phi(X) = AX - X \pmatrix{0& -a\\a & 0}. $$

We also find that because the block entries of $A^*$ commute, the inverse of $A^*$ can be found in a straightforward fashion.

Answer 2

Yes! The inverse is easy to find because each n-by-n block commute with each other:

$$A^{*-1} = (A^2 + a^2I)^{-1}*\begin{pmatrix} A & aI \\ -aI & A \end{pmatrix}$$

Here, the $*$ denotes matrix multiplication with each of the n-by-n blocks.

As for the eigenvalues, we see that the characteristic polynomial of $A^*$ is $\det{((A-\lambda I)^2+a^2I)}$. From here we see eigenvalues of the form $\lambda \pm ia$ where $\lambda$ is an eigenvalue of $A$.