What are the eigenvalues of $\begin{bmatrix}0 & I\\\alpha I & D\end{bmatrix}$ where $D$ is diagonal?

Question

Is there a trick to calculate the eigenvalues of $\begin{bmatrix}0 & I\\\alpha I & D\end{bmatrix}$ where $D$ is diagonal? I'm looking for some kind of expression in terms of $\alpha$ and $D$ if one exists.

Answer 1

There is no need to use any trick in this problem, just do it in simple way. After simple calculation, there will be two eigenvalues which will come: one is Alpha and the other one is (Alpha+D). That's it!!

Answer 2

Well, it looks like it can just be ground out using the characteristic polynomial. Note, $$ p(\lambda) = \det\left(\begin{bmatrix}-\lambda I & I\\\alpha I & D-\lambda I\end{bmatrix}\right) = \det (\lambda^2 - \lambda D - \alpha I). $$ Since $D$ is diagonal, the determinant is simply the product of the diagonal. Hence, if $d$ is a diagonal element of $D$, the eigenvalues of the larger system are simply the roots of $\lambda^2 - d\lambda - \alpha$, which are

$$ \frac{d\pm\sqrt{d^2+4\alpha}}{2} $$

Answer 3

$$v_2 = \lambda v_1$$ $$\alpha v_1 + Dv_2= \lambda v_2$$

$$\alpha v_1 = (\lambda I -D)v_2=(\lambda I -D)\lambda v_1$$

$$((\alpha-\lambda^2 )I+\lambda D)v_1=0$$

$$\det(\alpha-\lambda^2 )I+\lambda D)=0$$

$$\prod_{i=1}^n(\alpha-\lambda^2 +\lambda d_i)=0$$