I wonder if there is any inequality (or even equality) relating $Trace(AB (AB)^T)$ and $Trace(AA^T)$, where $A, B$ are both general square matrices of dimension $p$, and $B$ is positive semidefinite (thus symmetric). I did not find much discussion on this, and I tried thinking of it in terms of trace as an inner product, but haven't made much progress yet. Any help or pointer to books is appreciated!
Asked
Active
Viewed 238 times
1 Answers
1
Yes, using the cyclic property of the trace, one has $$\operatorname{tr}(AB(AB)^T)=\operatorname{tr}(ABB^TA^T)=\operatorname{tr}(A^T A B B^T).$$ Since both $A^T A$ and $B B^T$ are positive semidefinite (try proving this), one can use the following inequality for positive semidefinite matrices $C,D$: $$\operatorname{tr}(CD)\leq \operatorname{tr}(C)\operatorname{tr}(D).$$ Therefore $$\operatorname{tr}(AB(AB)^T)\leq \operatorname{tr}(A^T A)\operatorname{tr}(BB^T)=\operatorname{tr}(AA^T)\operatorname{tr}(B^2).$$
projectilemotion
- 15,914