How to find the eigenvalues and eigenvectors of this block matrix?

Question

What is the best way to find eigenvalues and eigenvectors of the following matrix

\begin{bmatrix} -a\mathbf{I}& \mathbf{D}\\ \mathbf{I}& \mathbf{0}\end{bmatrix}

where $\mathbf{I}$ is of size $n \times n$, $\mathbf{D}$ is a diagonal matrix of size $n \times n$ and $a$ is a scalar?

Also, is there a name for this type of block matrix?

Answer 1

\begin{equation}\begin{bmatrix} -a\mathbf{I}& \mathbf{D}\\ \mathbf{I}& \mathbf{0}\end{bmatrix} \begin{bmatrix}\boldsymbol{v}_1 \\ \boldsymbol{v}_2\end{bmatrix}=\lambda\begin{bmatrix}\boldsymbol{v}_1 \\ \boldsymbol{v}_2\end{bmatrix} \end{equation} Where $\boldsymbol{v}_1$ and $\boldsymbol{v}_2$ are $n$ dimensional vectors. It follows: $-a v_1+D v_2 =\lambda v_1$ and $I v_1=\lambda v_2$. From the last equation, you can get directly two values for $\lambda$ and respectively one corresponding easy relations for $v_1$ and $v_2$

Answer 2

Let

$${\rm B} := \begin{bmatrix} -a \, {\rm I}_n & {\rm D}\\ {\rm I}_n & {\rm O}_n\end{bmatrix}$$

and let the characteristic polynomial of diagonal matrix ${\rm D} = \mbox{diag} (d_1, \dots, d_n)$ be

$$q_D (s) := \prod_{i=1}^n (s - d_i)$$

Hence, the characteristic polynomial of block matrix $\rm B$ is

$$\det \left( s \, {\rm I}_{2n} - {\rm B} \right) = \det \begin{bmatrix} (s+a) \, {\rm I}_n & -{\rm D}\\ -{\rm I}_n & s \, {\rm I}_n\end{bmatrix} = \det \left( s (s+a) \, {\rm I}_n - {\rm D}\right) = q_D \left( s (s+a) \right)$$

and, thus, the $2n$ eigenvalues of $\rm B$ are the roots of the $n$ quadratic polynomials

$$s (s+a) - d_i = s^2 + a s - d_i$$

The eigenvectors can be found by solving the corresponding linear systems.