# eigenvalues and eigenvectors of 2x2 block matrix

My question is a really straightforward one: Is there an easier way to find the eigenvalues and/or eigenvectors of a 2x2 block diagonal matrix other than direct diagonalization of the whole matrix?

$\left( \begin{array}{ccc} A & 0 \\ 0 & B \end{array} \right )$

Here $A$, $B$ are $n \times n$ symmetric matrices.

• Evidently, by computing separetely the eigenvalues of $A$ and $B$ because $det(M-\lambda I_{2n})=det(A-\lambda I_n)det(B-\lambda I_n)$ (determinant of a block diagonal matrix = product of the determinants of its blocks). The fact that $A$ and $B$ are symmetric is unimportant. – Jean Marie Apr 23 '16 at 16:32

Since $A$ and $B$ are symmetrical, we can diagonalize them in the way such that $A=UD_AU^T$ and $B=VD_BV^T$.
$\left(\begin{array}{cc} A & 0 \\ 0 & B \end{array} \right)=\left(\begin{array}{cc} UD_AU^T & 0 \\ 0 & VD_BV^T \end{array} \right)=\left(\begin{array}{cc} U & 0 \\ 0 & V \end{array} \right)\left(\begin{array}{cc} D_A & 0 \\ 0 & D_B \end{array} \right)\left(\begin{array}{cc} U & 0 \\ 0 & V \end{array} \right)^T.$