Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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|cite|improve this question
up vote 6 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|cite|improve this answer

I was about to ask the very same question when I found this one... I know that I'm two years late but this issue does intrigue me so I want to say that the formula in the question, although not valid in general, can be still useful in some situations to compute the eigenvalues of block matrix. In fact if $\lambda$ is an eigenvalue of $\begin{pmatrix} A & B\\ C & D\\ \end{pmatrix}$ then at least one of the following must be true:

  1. $\lambda$ is an eigenvalue of $A$;
  2. $\lambda$ is a solution of $\det(D - \lambda I - C(A-\lambda I)^{-1}B)=0$.

The proof of this fact is just the observation made in the question. I want to say that the condition 2 is NOT a polynomial in $\lambda$. I think that it's a rational function instead. Thus if it is wanted to compute the eigenvalues of the block matrix, one could just compute the roots of $\det(A - \lambda I)$, then the roots of the numerator of $\det(D - \lambda I - C(A-\lambda I)^{-1}B)$ and then manually check which one of those is an actual eigenvalue of the big matrix. Ok, I admit that it's not very efficient but perhaps in some special cases it could be useful.

share|cite|improve this answer

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.