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.

  • 1
    $\begingroup$ Have you tried the singular value decomposition of $B$? $\endgroup$ – user251257 Jun 25 '18 at 19:35
  • $\begingroup$ @user251257 : no $\endgroup$ – 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 $$A \pmatrix{u\cr -v\cr} = \pmatrix{-Bv\cr B^T u\cr} = - \lambda \pmatrix{u\cr -v\cr}$$


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