Determining the eigenvalues of a $4×4$ block diagonal matrix.

Question

I was wondering if I found my eigenvalues correctly for the following matrix:

$$A = \begin{bmatrix} -14 & 8 &0&0 \\ -12 &6&0&0 \\ 0&0&-20&21 \\ 0&0&-28&29 \end{bmatrix}$$

$\lambda = -6, -2, 1, 8.$

How I found the eigenvalues is by splitting this $4\times4$ matrix into $2\times2$ block matrices and finding their eigenvalues for those matrices. It seemed to work, but is this a correct way to go about it?

Answer 1

For every block matrix

$$A = \begin{pmatrix} B & 0 \\ 0 & C \end{pmatrix} $$

with $B, C \in \mathbb{C}^{n \times n}$ it holds that $\det(A) = \det(B) \det(C)$.

Every eigenvalue of your matrix $A$ satisfies $\det(A - \lambda I ) = 0$. Hence such an eigenvalue also satisfies

$$\det(A - \lambda I) = \det(B - \lambda I) \det(C - \lambda I) = 0$$

and your reasoning is correct.

Answer 2

Suppose $\lambda$ is an eigenvalue of the upper-left $2×2$ submatrix of $A$ with corresponding non-zero eigenvector $x$. Let $y=\begin{bmatrix}x\\0\\0\end{bmatrix}$. Then, since the two added zeros remain zero upon multiplication, $$Ay=\lambda y$$ Similar reasoning holds for the lower-right block. Hence your reasoning is correct.

Answer 3

The eigenvalues of $A$ are values of $λ$ that satisfy the equation $|A- \lambda I|=0$. So what you should do to found the eigenvalues is: $$|A- \lambda I|= \left| \begin{matrix} -14-\lambda & 8 & 0 & 0\\ -12 & 6-\lambda & 0 & 0 \\ 0 & 0 & -20 -\lambda & 21 \\ 0 & 0 & -28 & 29-\lambda \end{matrix}\right| = (-14-\lambda)\left| \begin{matrix} 6-\lambda & 0 & 0 \\ 0 & -20 -\lambda & 21 \\ 0 & -28 & 29-\lambda \end{matrix}\right|+12 \left| \begin{matrix} 8 & 0 & 0\\ 0 & -20 -\lambda & 21 \\ 0 & -28 & 29-\lambda \end{matrix}\right|=λ^4-λ^3-52λ^2-44λ+96=0 $$ Then, you have $λ=1,\:λ=8,\:λ=-2,\:λ=-6$ as eigenvalues. Indeed, "splitting" this $4\times4$ determinant into two $2\times2$ determinants is right way to get the characteristic polynomial of the matrix.