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.

  • $\begingroup$ 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 '16 at 16:32

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


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

  • $\begingroup$ 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 '16 at 16:53
  • $\begingroup$ Thanks! That's exactly what I was looking for! $\endgroup$ – Sr Incerteza Apr 23 '16 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.