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Let $A$ be an $n\times n$ positive definite matrix with eigenvalues $a_1,a_2,\dots,a_n$ in descending order. Let $T$ be an $n\times n$ positive definite matrix with eigenvalues in $t_1,\dots,t_n$ in descending order, and consider the matrix $B=(A+T)^{-1}$.

My intuition is that the trace of $AB$ should be less than $n$. Or that the spectral norm of $AB$ should be less than $1$.

The intuition is that if $A$ and $T$ where mutually diagonalizable, then one could express the trace as $$\sum_{i=1}^n \frac{a_i}{a_i+t_i},$$ which is plainly less than $n$.

But of course $A$ and $T$ are not mutually diagonalizable. Is there a counterexample if they are not, or will the trace always be less than $n$?

EDIT: I will leave this question up for now, but I think I figured it out. You can use the Weyl inequalities to get bounds on each eigenvalue of $(A+T){-1}$. Then you can use Von Neumann's trace inequality to finish the proof.

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  • $\begingroup$ Actually I think this might follow from the Weyl inequalities. I am trying to work it out. $\endgroup$ – Lepidopterist Dec 20 '15 at 22:21
  • $\begingroup$ It may be useful to note that the matrix $A^{1/2}(A + T)^{-1}A^{1/2}$ is similar to $AB$ and is positive definite. $\endgroup$ – Ben Grossmann Dec 20 '15 at 22:41
  • $\begingroup$ Do you not agree that what I wrote is sufficient? By the Weyl inequalities, the ith eigenvalue of $(A+T)^{-1}$ is at most $\frac{1}{\lambda_i(A)+\lambda_n(T)}$. Then you can use Von Neumann's trace inequality. And as to your question earlier about the trace, our discussion was erased, but if you did find a way to bound the spectral norm by 1, then the trace can at most be $n$ times that since the trace is the sum of the eigenvalues. $\endgroup$ – Lepidopterist Dec 20 '15 at 22:45
  • $\begingroup$ Oh, I didn't remember what Weyl's says. Yes, your method works then. I'm not actually sure whether we have $\|AB\| \leq 1$ now, but we definitely have $\|A^{1/2}BA^{1/2}\| \leq 1$. $\endgroup$ – Ben Grossmann Dec 20 '15 at 22:48
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The trace of a product of matrices is invariant under cyclic permutation of the multiplicands (provided the permuted product is well-formed). So, we have $$ \operatorname{tr}A(A+T)^{-1} =\operatorname{tr}A^{1/2}(A+T)^{-1}A^{1/2} =\operatorname{tr}(I+A^{-1/2}TA^{-1/2})^{-1}. $$ Since $A^{-1/2}TA^{-1/2}$ is positive definite, all eigenvalues of $I+A^{-1/2}TA^{-1/2}$ are strictly greatly than $1$ and the result about the trace follows.

As for the spectral norm (remark: when the matrix is not Hermitian, the term "spectral norm" might be misleading; it'd better be called an "operator norm"), no, $\|A(A+T)^{-1}\|$ is not necessarily smaller than $1$ (although its eigenvalues are always all smaller than $1$). Here is a random counterexample generated by computer: $$ A=\pmatrix{6&-5\\ -5&6}, \ T=\pmatrix{3&-5\\ -5&9}, \ \|A(A+T)^{-1}\|\approx 1.2751. $$

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  • $\begingroup$ Are you sure about your counterexample? I just ran it and got the largest eigenvalue is .9129. I believe the proof I gave in my question works. $\endgroup$ – Lepidopterist Dec 21 '15 at 11:25
  • $\begingroup$ @Lepidopterist Yes, the largest-sized eigenvalue of $AB$ is indeed 0.9129, but the norm is 1.2751. That's why I remarked that it is misleading to call the norm "spectral norm" when the matrix is not Hermitian. The so-called spectral norm is actually a function of singular values, not eigenvalues. $\endgroup$ – user1551 Dec 21 '15 at 13:00
  • $\begingroup$ Ah, I understand now. Thanks for the correction. $\endgroup$ – Lepidopterist Dec 21 '15 at 15:01

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