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!



Eigenvalues of a block matrix with zero diagonal blocks
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 $\epsilon\rightarrow 0$.


