This question has been in asked in a few varieties here but not in this one.
If we have a real, symmetric, positive-definite matrix $A$ and a real, symmetric, positive-definite matrix $B$ and we know an eigenvalue of each of them say, $l$ and $k$ respectively, can we express the eigenvalues of the product $AB$ in terms of $l$ and $k$?
Moreover, through some helpful comments, I have learned that they need to be simultaneously diagonalizable. So suppose they're inside the trace, so $tr(AB)=tr(BA)$, does that admit an expression that involves only the eigenvalue of one of them?
Thank you so much!
Best Answer
Let $\Lambda_A$ and $\Lambda_B$ be the spectra of $A$ and $B$, respectively. If $\mathrm{tr}(AB)$ was dependent only on the spectra of $A$ and $B$, that is, $$ \mathrm{tr}(AB)=f(\Lambda_A,\Lambda_B), $$ then for any orthogonal matrices $U$ and $V$, $\tilde{A}:=U^TAU$ and $\tilde{B}:=V^TBV$ would be still SPD with the spectra $\Lambda_A$ and $\Lambda_B$ and $\mathrm{tr}(\tilde{A}\tilde{B})=f(\Lambda_A,\Lambda_B)=\mathrm{tr}(AB)$. This is simply not true.
Consider $$ A:=\pmatrix{1&0\\0&2}, \quad B:=\pmatrix{2&0\\0&1}, \quad \mathrm{tr}(AB)=4. $$ Let $$ U:=\frac{1}{\sqrt{2}}\pmatrix{1&1\\-1&1}, \quad V:=\pmatrix{0&1\\-1&0} $$ be two orthogonal matrices and set $$ \tilde{A}=U^TAU=\frac{1}{2}\pmatrix{3&-1\\-1&3},\quad \tilde{B}=V^TBV=\pmatrix{1&0\\0&2}. $$ Clearly, $\mathrm{tr}(\tilde{A}\tilde{B})=\frac{9}{2}\neq 4=\mathrm{tr}(AB)$, even though both $A$ and $\tilde{A}$ (and $B$ and $\tilde{B}$) have the same eigenvalues. Consequently, $\mathrm{tr}(AB)$ is not a function depending only on the eigenvalues of $A$ and $B$.