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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?

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@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

2 Answers 2

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.

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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.

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