$\begingroup$Let $A$ and $B$ be two arbitrary matrix with proper dimension for multiplication.
Consider this trace inequlaty which is trace of multiplication of two matrices versus their individual traces
$$\text{tr}(AB) \leq \text{tr(A)} \text{tr(B)}$$
1- Do we have result for rectangular matrix that satisfy this inequality?
2- If they were square matrices what are the conditions?
3- Is there any specific name for this inequality?
$\endgroup$41 Answer
$\begingroup$Can't think of anything deep, but if both $A$ and $B$ are positive semidefinite, the inequality is true: when $a=\operatorname{tr}(A)$, we have $A\preceq aI$ and hence $$ \operatorname{tr}(AB) =\operatorname{tr}(B^{1/2}AB^{1/2}) \le\operatorname{tr}(B^{1/2}(aI)B^{1/2}) =\operatorname{tr}(A)\operatorname{tr}(B). $$ This also follows from (and hence is weaker than) von Neumann's trace inequality, which in this context says that $$ \operatorname{tr}(AB)\le\sum_i\lambda_i(A)\lambda_i(B) $$ when the eigenvalues of $A$ and $B$ are arranged in the same (ascending or descending) order.
$\endgroup$8
