Is inequality $tr(A^{-1^T} B) tr(A^T B^{-1}) \leq constant$ correct?

Question

I have the following optimization problem \begin{align} \min_{A} &tr(A^{-1^T} B)\cr \text{subject to} &x^T A x > 0 \cr & A_{ii}=1 \end{align} where $A$ and $B$ are some positive definite matrices. Is the above problem is equivalent to the following problem: \begin{align} \max_{A} &trace(A^T B^{-1})\cr \text{subject to} &x^T A x > 0\cr & A_{ii}=1 \end{align} In other words is this inequality $ tr(A^{-1^T} B) tr(A^T B^{-1}) \leq c$ where $c$ is some constant, is correct? And finaly what if $A$ and $B$ are positive semidefinite and we replace inverse with pesudo inverse?

Answer 1

Set $X={A^{-1}}^T B$ where I assume $A$ and $B$ are $n\times n$ positive definite matrices. Then, $X$ has eigenvalues $\lambda_1,\ldots,\lambda_n>0$, and the trace $\text{tr}(X)=\sum_j\lambda_j$.

The constraint $A_{ii}=1$ (added after my original answer) have very little impact on the eigenvalues, and so I'll ignore them.

Since $\text{tr}(UV)=\text{tr}(VU)$ for all matrices $U$ and $V$ (of appropriate dimensions), $$ \text{tr}(A^TB^{-1})=\text{tr}(X^{-1})=\sum_j\frac{1}{\lambda_j}. $$ Thus, the product of the traces becomes $$ \text{tr}(X)\cdot\text{tr}(X^{-1}) =\sum_i\lambda_i\cdot\sum_j\frac{1}{\lambda_j} \ge n^2. $$

To see that there is no upper bound, for example let $X$ be a diagonal $2\times2$ matrix with eigenvalues $1$ and $u$. Then, the product of the traces becomes $(1+u)(1+1/u)=2+u+1/u$ which can get arbitrary large as $u\rightarrow0^+$ or $u\rightarrow\infty$.

If we require that $A_{ii}=1$, we can still force large variations in the eigenvalues: e.g., $A=\pmatrix{1&u\\ u&1}$ with $u>1$.

In order to obtain an upper bound, it would be necessary to place constraints that limit the ratio between the largest and smallest eigenvalues. An upper bound on the eigenvalues of $X$ could for example be added by demanding that $\text{tr}(X)\le C$ for some fixed $C$; or $|X|^2=\text{tr}(X^TX)\le C$. Lower bounds on the eigenvalues may be placing similar restrictions on $X^{-1}$.

Instead of placing the bounds on $X$ and $X^{-1}$, I think it should be sufficient to place similar bound on $A$, $A^{-1}$, $B$, and $B^{-1}$.