Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Based on the next relation:

$$\det\begin{bmatrix}A & B \\ C & D\end{bmatrix} = \det(A)\det(D - CA^{-1}B),$$

I have that for computing the eigenvalues of the block matrix:

$$\det\begin{bmatrix}A-\lambda I & B \\ C & D-\lambda I\end{bmatrix} = \det(A-\lambda I)\det((D-\lambda I) - C(A-\lambda I)^{-1}B) = 0$$

So $\det(A - \lambda I) = 0$ says that the eigenvalues of $A$ are eigenvalues of the block matrix? But from some numerical simulations I have found that this is not true, what am I missing here? Maybe is because the first relation requires $A$ nonsingular and $A-\lambda I$ is not?

Then, this leads me to another question, why this expression holds $$\det\begin{bmatrix}A-\lambda I & 0 \\ C & D-\lambda I\end{bmatrix} = \det(A-\lambda I)\det((D-\lambda I) = 0$$ for stating that the eigenvalues of the block matrix are the eigenvalues of $A$ and $D$ if $A-\lambda I$ is singular?

Many thanks in advance.

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

If $\det(A - \lambda I) = 0$, then you cannot form $(A - \lambda I)^{-1}$, which appears also in the formula. Hence the formula does not apply, unless $B = 0$ or $C = 0$, the case of a block tridiagonal matrix.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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