Eigenvalues of an off-diagonal block symmetric matrix

Question

Consider the the following symmetric off-diagonal block matrix:

$$A = \begin{bmatrix} 0 & B \\ B' & 0\end{bmatrix}$$

where $B$ is rectangular. How are the eigenvalues of $A$ related to the eigenvalues of $BB'$?

For the case where $B$ is square, there is a solution here, namely the eigenvalues of $A$ are the square roots of the eigenvalues of $B'B$.

How can I deal with the case where $B$ is rectangular?

By the square matrix $$A^2 = ((BB',0),(0,B'B) $$ is a diagonal block matrix of square matrices. — Roland F, May 03 '24 at 18:50

score 2 · Accepted Answer · answered May 03 '24 at 18:54

2

You use the Schur complement for block matrices. Suppose $B$ is $n\times m$ and wlog $n \geq m$, then $$\det(\lambda I -M)= \lambda^n \det(\lambda I - \lambda^{-1}B’B) = \lambda^{n-m}\det(\lambda^2 I - B’B).$$

If $B$ is rectangular it’s not that different, you get the square roots of the eigenvalues of $B’B$ and then some zero eigenvalues to make up for the missing ranks.

answered May 03 '24 at 18:54

Lee Fisher

2,749

[+1] A detail : adjective "some" isn't adequate because $n-m$ can be equal to an arbitrarily large value... – Jean Marie May 03 '24 at 19:02
Thank you. How do you deal with the case $\lambda=0$? – Alphie May 03 '24 at 20:31
Im not sure I understand what you mean by “deal with”. We have that $0$ is an eigenvalue with multiplicity at least n-m. – Lee Fisher May 04 '24 at 03:04

Eigenvalues of an off-diagonal block symmetric matrix

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