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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.

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  • $\begingroup$ 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. $\endgroup$
    – Jean Marie
    Apr 23, 2016 at 16:32

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Since $A$ and $B$ are symmetrical, we can diagonalize them in the way such that $A=UD_AU^T$ and $B=VD_BV^T$.

Hence,

$\left(\begin{array}{cc} A & 0 \\ 0 & B \end{array} \right)=\left(\begin{array}{cc} UD_AU^T & 0 \\ 0 & VD_BV^T \end{array} \right)=\left(\begin{array}{cc} U & 0 \\ 0 & V \end{array} \right)\left(\begin{array}{cc} D_A & 0 \\ 0 & D_B \end{array} \right)\left(\begin{array}{cc} U & 0 \\ 0 & V \end{array} \right)^T.$

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  • $\begingroup$ Remark: I agree that symmetric is not important if diagonalization of each matrices exists. the only change is that we just have to change the tranpose to inverse. Symmetric matrices promise us that they can be diagonalize. $\endgroup$
    – Siong Thye Goh
    Apr 23, 2016 at 16:53
  • $\begingroup$ Thanks! That's exactly what I was looking for! $\endgroup$
    – Sr Incerteza
    Apr 23, 2016 at 16:53

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