On what condition $trace(A) \ge trace(AB)$?

Question

Considering $A$ and $B$ are positive semidefinite real symmetric matrices, on what conditions we can have $trace(A) \ge trace(AB)$?

Answer 1

Under the extra assumption that $A,B$ commute: Then $A,B$ are orthogonally diagonalizable simultaneously.

Let $P$ be an orthogonal matrix, and $D_1,D_2$ be diagonal such that $$A=PD_1P^T \\ B=PD_2P^{T}$$

Then $$AB=PD_1D_2P^{T}$$

Therefore, if $\lambda_1,.., \lambda_n$ are the eigenvalues of $A$ and $\beta_1,.., \beta_n$ are the eigenvalues of $B$, in any order, there exists a permutation $\sigma$ such that $$\tr(AB)=\lambda_1 \beta_{\sigma(1)}+...+\lambda_n \beta_{\sigma(n)}$$

The question you are asking in this case is basically the following

Question Given some non-negative numbers $\lambda_1,.., \lambda_n$ and $\beta_1,.., \beta_n$, (and a permutation $\sigma$ which can actually be ignored by reordering $\beta$) under which condition do we have $$\lambda_1 \beta_{\sigma(1)}+...+\lambda_n \beta_{\sigma(n)}\leq \lambda_1+..+\lambda_n ?$$

Note that for any such $\lambda, \beta$'s you can take $A,B$ to be the diagonal matrix with those numbers on the diagonal for an example/counterexample.

If $A$ is not the zero matrix, then the most obvious answer is : all eigenvalues of $B$ are $\leq 1$.

If $A,B$ don't commute, the question seems to be way to complicated.