2
$\begingroup$

I have a matrix whose determinant is zero:

$$\det\begin{bmatrix}A-I\lambda&B\\C &D-I\lambda \end{bmatrix} = 0$$

where $\lambda$ is a vector of complex scalars, I is an identity matrix, and A,B,C,D are square, complex, and invertible.

From block matrix identities using the Schur compliment, we know that we can then write

$$\det(A-I\lambda) \cdot det(D-I\lambda - C (A-I\lambda)^{-1} B) = 0$$

Does the above equation suggest that there are two solutions (sets of eigenvalues) $\lambda$? One being $eig(A)$ and the other being whatever $\det(D-I\lambda - C (A-I\lambda)^{-1} B)$ works out to be?

I've done some numerical tests, and it seems that the first determinant does not give the correct eigenvalues, which does not make sense to me.

What about the slightly modified case of

$$\det\begin{bmatrix}A-I\lambda_x&B\\C &D-I\lambda_y \end{bmatrix} = 0$$

in which we can say

$$\det(A-I\lambda_x) \cdot \det(D-I\lambda_y - C (A-I\lambda_x)^{-1} B) = \det(D-I\lambda_y) \cdot \det(A-I\lambda_x - B (D-I\lambda_y)^{-1} C) = 0$$

Can we then say that simply $\lambda_x = eig(A)$ and $\lambda_y = eig(D)$? This seems fundamentally incorrect to me, but I'm not sure as to why.

$\endgroup$
  • 2
    $\begingroup$ To use the Schur identity, you need $A-\lambda I$ to be invertible, hence $\lambda $ cannot be an eigenvalue of $A$ in this case. $\endgroup$ – copper.hat Apr 19 '16 at 20:14
  • $\begingroup$ Thanks for pointing that out - I wasn't aware that $A-\lambda I$ wasn't invertible. That would explain why it wasn't correct for me. Can you then say the same for $A-\lambda_x I$? $\endgroup$ – Stuart Barth Apr 19 '16 at 20:30
  • $\begingroup$ Well, to use the Schur identity, you need the top left block (or bottom right) to be invertible, so in the latter case, $\lambda_x$ cannot be an eigenvalue of $A$. $\endgroup$ – copper.hat Apr 19 '16 at 20:43
0
$\begingroup$

@copper.hat 's comment was the correct answer -- since det$(A-I\lambda)$ and det$(D-I\lambda)$ must be invertible, they must have a non-zero determinant.

Therefore, they cannot possibly be the solution to the eigenmode equation.

$\endgroup$

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.