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: \begin{equation} \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) \end{equation}

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.

  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Aug 2 '15 at 19:14

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.


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