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Suppose $A$ is a $k_1\times k_2$ matrix with real entries, $k_1<k_2$. 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!

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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 $\epsilon\rightarrow 0$.

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  • $\begingroup$ I see. What about the first determinant there, wouldn't that have a root of multiplicity k2-k1? $\endgroup$ – AdamNie May 2 at 16:44
  • $\begingroup$ this root would cancel with the pole of the inverse $X^{-1}$. $\endgroup$ – Carlo Beenakker May 2 at 17:43
  • $\begingroup$ 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? $\endgroup$ – AdamNie May 2 at 17:56
  • $\begingroup$ 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. $\endgroup$ – Carlo Beenakker May 2 at 19:06

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