Eigenvalues of Certain Symmetric Block Matrix

What can we say about the relation between the eigenvalues of the following block matrix with identity diagonal blocks, and the singular values of the off-diagonal blocks: $$\Gamma=\left( {\begin{array}{cc} I_{k\times k} & B_{k\times n-k} \\ B_{n-k\times k}^{T} & I_{n-k\times n-k} \\ \end{array} } \right)$$

We know (also from my other questions) that if the blocks are all of the same size (\frac{n}{2}) then we have: $$\lambda_{i}(\Gamma)^{\pm}=1\pm\sigma_{i}(B)$$. What about the general case where the sizes are not equal. Consider for the sake of clarity $0\leq k\leq \frac{n}{2}$. I believe $k$ of them are equal to unity.

• The answer will ultimately be "kind of the same, but with extra zeros depending on how you define singular values". So, here's a question: if $B$ is $m \times n$, then how many singular values does $B$ have when $m < n$? What about if $m > n$? With the definition that I'm used to, $B$ always has $n$ singular values. – Omnomnomnom Aug 2 '15 at 19:14

1 Answer

Here's a helpful start. Suppose first that $k < n/2$. The SVD then gives us $$B = U\pmatrix{\Sigma & 0}V^*$$ where $\Sigma$ is square an diagonal, and $0$ is a block matrix of $0$s.

Define $$\tilde U = \pmatrix{U\\& V}$$ We then have $$\tilde U^* \Gamma \tilde U = \pmatrix{ I & \Sigma & 0\\ \Sigma & I & 0\\ 0 & 0 & I}$$ As you can verify by block-matrix multiplication.

Similarly, when $k > n/2$ and defining $\tilde U$ as above, you end up with $$\tilde U^* \Gamma \tilde U = \pmatrix{ I & 0 & \Sigma \\ 0 & I & 0\\ \Sigma & 0 & I}$$ In either case, the answer is something like "you get a few extra $1$s as eigenvalues".

My full answer for your case would be as follows: when $k < n/2$, $B$ has $k$ singular values. We then get $n - 2k$ additional eigenvalues of $\Gamma$ in addition to those determined by the singular values, and each additional eigenvalue is unity.