# Properties of specific matrix

In my studies of linear algebra and its applications I have come across the following type of symmetric block matrix:

$A = \begin{pmatrix} 0_{n \times n} & B \\ B^T & 0_{m \times m} \end{pmatrix} \quad ; \quad m \leq n$

Where $B$ is a $n \times m$ matrix.

I would like to know the following about the matrix:

1. The inertia of $A$ (the number of positive, negative and zero eigenvalues) as a function of the rank of $B$.

2. Eigenvalue bounds (or the eigenvalues themselves) of $A$.

I noticed some pattern with task 1 as for $B$ of rank 1 all eigenvalues are zero except for two nonzero ones of opposite signs with equal absolute values, and with $B$ of rank 2 all except 4 are zero, with two pairs of nonzero eigenvalues each of opposite signs and equal absolute values, and so forth. I was wondering if someone could help me make this formal and prove it? I thought about Sylvester's law of inertia for matrices but I cannot apply it here.

Regarding eigenvalue bounds, I was wondering if there exist closed-form formulas? Maybe non-trivial approximations on the behavior of the eigenvalues? I have no idea here. All help is appreciated.

• Have you tried the singular value decomposition of $B$? – user251257 Jun 25 '18 at 19:35
• @user251257 : no – kroner Jun 25 '18 at 19:58

By the block matrix determinant formula, $$\det(A - \lambda I) = (-\lambda)^n \det(-\lambda I_m + \lambda^{-1} B^T B) = \lambda^{n-m}\det(B^T B - \lambda^2 I_m)$$ (the middle expression is only defined for $\lambda \ne 0$, but taking limits we find that the first and last are equal for all $\lambda$).
Thus the eigenvalues are $\pm$ the square roots of the eigenvalues of $B^T B$, with an additional $n-m$ eigenvalues of $0$.
The fact that the nonzero eigenvalues come in $(+,-)$ pairs can also be seen from the fact that if $A \pmatrix{u\cr v\cr} = \lambda \pmatrix{u \cr v\cr}$, i.e. $Bv = \lambda u$ and $B^T u = \lambda v$, then $$A \pmatrix{u\cr -v\cr} = \pmatrix{-Bv\cr B^T u\cr} = - \lambda \pmatrix{u\cr -v\cr}$$