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.

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

I do know generally $\text{trace}(A^{-1}B)\not= \sum_i \lambda_{B_i}/\lambda_{A_i}$,

where $\lambda_{A_i}$ and $\lambda_{B_i}$ are the corresponding eigenvalues of matrix $A$ and $B$ respectively,

but is there any cases when this equality can be statified?

share|cite|improve this question
@sbr for $A=B$ you need to know which is the first eigenvalue and which is the second and so on, else the equality is not true in general – Dominic Michaelis Mar 8 '13 at 8:30
up vote 1 down vote accepted

If $A$ and $B$ are simultaneously diagonalisable (or trigonalisable) and you are combining eigenvalues for the same (generalised) eigenvectors then this obviously holds. In other cases (i.e., almost always), all bets are off.

share|cite|improve this answer

There are many cases: let $A$ be $\gamma \cdot I$ where $I$ is the identiy matrix and $\gamma$ is an arbitrary scalar $\neq 0$, or let $B=0$ (the matrix with every entry zero).

If not every eigenvalue is the same you should give an order of them.

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.