# Finding eigenvalues of block 2x2 matrix

In my studies of probability theory, I have recently come across the following eigenvalue problem as part of my research:

Let $\Sigma$ be a given positive definite $d \times d$ matrix, and let $p > 0$ be a positive constant, we look at the following block matrix

$M = \left[ {\begin{array}{cc} \Sigma + pI_d & \Sigma \\ \Sigma & \Sigma \\ \end{array} } \right]$

Note: $I_d$ is the d-dimensional identity matrix.

My question here: How can I compute the eigenvalues of the block matrix $M$ supposing only that we know eigenvalues and eigenvectors of $\Sigma$.

I have tried using known results, and yet the eigenvalues elude me. I thought about using Schur complements, but can't figure that out. Also, I thought maybe a brute force solution would work, yet that failed also. I cannot seem to figure out the eigenvalues, so I would certainly appreciate any help on this. I thank all helpers.

• About Schur complement: You can calculate that by Gaussian elimination on $M$. Since Gaussian elimination does not preserve eigenvalues, I doubt that this will lead to the eigenvalues of $M$. Correct me, if that is wrong. – P. Siehr Aug 22 '17 at 7:22
• @P.Siehr : I agree, thanks for the comment – kroner Aug 22 '17 at 7:22
• I can't answer that question, but maybe this will help someone else: If you plug in $v:=(v_1,v_2)^t$, two eigenvector of $Σ$, then you get the condition $$2λ_1-p\overset{!}{=}2λ_2.$$ So if two eigenvalues fulfil that condition, $v$ is an eigenvector with eigenvalue $2λ_2$. – P. Siehr Aug 22 '17 at 7:31
• @P.Siehr : Thanks for a lovely comment – kroner Aug 22 '17 at 7:33
• @P.Siehr : Very helpful perspective, I appreciate your idea – kroner Aug 22 '17 at 7:33

Since the blocks commute, the block matrix determinant formula says the characteristic polynomial of your matrix is \eqalign{ \det(M - \lambda I_{2d}) &= \det\pmatrix{\Sigma + (p-\lambda) I_d & \Sigma\cr \Sigma & \Sigma - \lambda I_d}\cr &= \det((\Sigma + (p-\lambda I_d)) (\Sigma - \lambda I_d) - \Sigma^2)\cr&= \det((p-2\lambda) \Sigma + \lambda (\lambda-p) I_d)} This is $0$ iff $\lambda(p-\lambda)/(p-2\lambda)$ is an eigenvalue of $\Sigma$. That is, corresponding to each eigenvalue $\mu$ of $\Sigma$, we have eigenvalues $\lambda = \mu + p/2 \pm \sqrt{\mu^2 + p^2/4}$ of $M$.