Eigenvalues of a matrix based on eigenvalues of its submatrices

Question

Let $A_1,A_2\in\mathbb{R}^{n\times n}$, and define
\begin{equation} A\triangleq\left[\begin{array}{cc}A_1&A_2\\I_n&0_n\end{array}\right]\in\mathbb{R}^{2n\times 2n}. \end{equation} Is there a way to express the eigenvalues of $A$ in terms of eigenvalues of $A_1$ and $A_2$?

The following equations hold: \begin{align} {\rm det~} (\lambda I_{2n}-A)&={\rm det~}\left(\lambda^2I_n-\lambda A_1-A_2\right),\\ {\rm det~} (\lambda I_{2n}-A)&={\rm det~}\left(\lambda I_n-A_1\right){\rm det~}\left(\lambda I_n-(\lambda I_n-A_1)^{-1}A_2\right), \end{align} but I am not sure if these equations are helpful.

Answer 1

Instead of trying to solve the determinant, I would try to play with eigenvalues/eigenvectors of $A_1$ and $A_2$.

If we have $A_1v_1=\lambda_1v_1$ and $A_2v_2=\lambda_2v_2$

Then for $v=\begin{pmatrix}v_1 \\ 0\end{pmatrix}$ we have $Av=\begin{pmatrix}A_1v_1 \\ v_1\end{pmatrix}=\begin{pmatrix}\lambda_1v_1 \\ v_1\end{pmatrix}=\lambda v\implies \begin{cases}\lambda_1v_1=\lambda v_1\\v_1=0\end{cases}$

This is not possible since $v_1\neq 0$.

For $v=\begin{pmatrix}0 \\ v_2\end{pmatrix}$ we have $Av=\begin{pmatrix}A_2v_2 \\ 0\end{pmatrix}=\begin{pmatrix}\lambda_2v_2 \\ 0\end{pmatrix}=\lambda v\implies \begin{cases}\lambda_2v_2=0\end{cases}$

May be possible if $0$ is eigenvalue of $A_2$, in that case $0$ is also eigenvalue of $A$.

For $v=\begin{pmatrix}av_1 \\ bv_2\end{pmatrix}$ we have $Av=\begin{pmatrix}A_1av_1+A_2bv_2 \\ av_1\end{pmatrix}=\begin{pmatrix}a\lambda_1v_1+b\lambda_2v_2 \\ av_1\end{pmatrix}=\lambda v$

$\implies \begin{cases}a\lambda_1v_1+b\lambda_2v_2=a\lambda v_1 \\ av_1=\lambda bv_2 \end{cases}$

May be possible if $A_1,A_2$ have common eigenvectors, then these are also eigenvectors for $A$ and $\lambda^2-\lambda_1\lambda-\lambda_2=0$. (at most two values given by solving the quadritic equation).

• One interesting case might be $A_1=A_2$ in which case for $\lambda=\lambda_1\big(\frac{1\pm\sqrt 5}{2}\big)$ then $v=\begin{pmatrix} \lambda v_1\\v_1\end{pmatrix}$ is an eigenvector for $A$ in which case we seem to have full $2n$ eigenvectors from the $n$ eigenvectors of $A_1$.

• Another interesting case might be $A_2$ nilpotent, and we are able to find half $n$ eigenvectors of $A$.

Except for these, which are partial results anyway (it covers certain eigenvalues and eigenvectors only) I don't see how to exploit further the characteristic elements of $A_1,A_2$ ? In general case there may not be any relation between eigenvalues of $A$ and these of $A_1,A_2$.

Answer 2

Except for the relation

$0$ is an eigenvalue of $A$ iff $0$ is an eigenvalue of $A_2$

nothing more can be said without knowing further geometry of the submatrices.

Example: take two matrices with eigenvalues being fixed to $1$ $$ A_1=\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix},\quad A_2=\begin{bmatrix}1 & a\\0 & 1\end{bmatrix} $$ and look at your first identity $$ \det(\lambda I-A)=\det\begin{bmatrix}\lambda^2-\lambda-1 & -a\\-\lambda & \lambda^2-\lambda-1\end{bmatrix}=(\lambda^2-\lambda-1)^2-\lambda a. $$ For any $\lambda\ne 0$ one can set $$ a=\frac{(\lambda^2-\lambda-1)^2}{\lambda}, $$ so basically nonzero eigenvalues of $A$ can be any, independent on the eigenvalues of $A_1$, $A_2$.