# Eigenvalues of a block matrix with zero diagonal blocks

Suppose $$A$$ is a $$k_1\times k_2$$ matrix with real entries, $$k_1. Let $$M$$ be the matrix $$\begin{equation} M:=\begin{pmatrix} 0_{k_1} & A\\ A^\top & 0_{k_2} \end{pmatrix}, \end{equation}$$ where $$0_k$$ denotes the $$k\times k$$ zero matrix. I know that if $$\lambda$$ is an eigenvalue of $$M$$ then $$\lambda^2$$ must be an eigenvalue of $$A^\top A$$. Since $$k_2>k_1$$, we can immediately conclude that $$M$$ has at least $$k_2 - k_1$$ zero eigenvalues.

I wish to obtain a generalization of this observation in the following sense. Suppose $$A_{12},A_{13}$$ and $$A_{23}$$ are $$k_1\times k_2$$, $$k_1\times k_3$$ and $$k_2\times k_3$$ dimensional matrices respectively and let $$\begin{equation} M:=\begin{pmatrix} 0_{k_1} & A_{12} & A_{13} \\ A_{12}^\top& 0_{k_2} & A_{23} \\ A_{13}^\top& A_{23}^\top& 0_{k_3} \end{pmatrix}. \end{equation}$$ My conjecture is that if $$k_3>k_1+k_2$$, then $$M$$ contains at least $$k_3-k_1-k_2$$ zero eigenvalues. I can't figure out how to prove it - any help/hint is appreciated!

## 1 Answer

If you decompose $$M=\begin{pmatrix} X_{q\times q}&Y_{q\times k_3}\\ (Y_{q\times k_3})^{\rm T}&0_{k_3\times k_3}\end{pmatrix}$$ into four block matrices, with $$q=k_1+k_2$$, then the determinant equals $$\det M=(-1)^{k_3}(\det X_{q\times q})\det[(Y_{q\times k_3})^{\rm T}X_{q\times q}^{-1}Y_{q\times k_3}].$$ The second determinant has a root of multiplicity $$k_3-q=k_3-k_1-k_2$$.

For $$k_1\neq k_2$$ the matrix $$X$$ is not invertible: We can give it an infinitesimal perturbation, $$M\mapsto M_\epsilon=M+\epsilon 1_{q\times q}$$, and then $$\det (\lambda-M_\epsilon)=\lambda^{k_3-q}f_\epsilon(\lambda)$$. The continuity of the determinant in the matrix elements ensures that the multiplicity of the root 0 cannot decrease in the limit $$\epsilon\rightarrow 0$$.

• I see. What about the first determinant there, wouldn't that have a root of multiplicity k2-k1? – AdamNie May 2 at 16:44
• this root would cancel with the pole of the inverse $X^{-1}$. – Carlo Beenakker May 2 at 17:43
• I think I understand what you meant now, thanks a lot. A further question: if in general I have $k_1<k_2<\ldots<k_n$ diagonal blocks of zero, without the assumption $k_n>\sum_{i\ne n}k_i$, can I say anything at all about the matrix? – AdamNie May 2 at 17:56
• My surmise is that you need one of the $k_i$'s, say $k_m$ to be larger than the sum of all the others, to have $k_m-\sum_{i\neq m}k_i$ zero eigenvalues. – Carlo Beenakker May 2 at 19:06