# trace of the inverse of a matrix times another matrix

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

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.
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).