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Suppose $A$ is a positive semi-definite matrix and we can bound its trace as $l \le tr(A) \le L$. I am wondering if it is possible to find the upper and lower bounds on the trace of $A^{-1}$ based on $tr(A)$ and its bound? What inequalities hold and under what conditions?

Edit: I should add that there is a lower bound for eigenvalues of the matrix $A$, i.e., $\lambda_i(A)\ge \lambda_0$.

Thanks

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  • $\begingroup$ It is possible to find a lower bound but not an upper bound. Just use the fact that the trace is the sum of eigenvalues, and that the inverse matrix has inverse eigenvalues. $\endgroup$
    – alesia
    Commented Jan 26, 2023 at 18:57
  • $\begingroup$ What if we know that there is a lower bound on eigenvalues? $\endgroup$
    – Amin
    Commented Jan 27, 2023 at 5:26

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Diagonalizing the positive semi-definite matrix $A$ (say $n\times n$) and assuming $0<l<L<\infty$, we reduce the problem to the following:

What are the best upper and lower bounds on $\sum_1^n \frac1{t_j}$ over all positive $t_j$'s such that $l\le\sum_1^n t_j\le L$?$

If $n\ge2$, we can take $t_1=(L+l)/2$ and $t_2,\dots,t_n$ close to $0$. So, we see that there is no finite upper bound on $\sum_1^n \frac1{t_j}$.

On the other hand, we have the inequalities $$\sum_1^n \frac1{t_j}\ge n^2\Big/\sum_1^n t_j\ge\frac{n^2}L, \tag{1}\label{1}$$ and the lower bound $\frac{n^2}L$ on $\sum_1^n \frac1{t_j}$ is exact, as it is attained when $t_j=L/n$ for all $j$; the first inequality in \eqref{1} follows immediately from (say) the HM-GM-AM-QM inequalities.

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  • $\begingroup$ Thanks for your response. I think this answer does not assume any lower bound on the eigenvalues of $A$. I am wondering if lower bound on eigenvalues has any impact? $\endgroup$
    – Amin
    Commented Jan 27, 2023 at 23:40
  • $\begingroup$ Any lower bound on the eigenvalues $t_j$ of $A$ must be $\le L/n$ for $tr(A)$ to be $\le L$. So, the values $t_j=L/n$ for all $j$ will be allowed, and these values provide for the exact lower bound $n^2/L$, given in the answer. So, the answer to your comment is: No, of course not -- any admissible lower bound on the eigenvalues will not have any impact. $\endgroup$
    – Iosif Pinelis
    Commented Jan 29, 2023 at 1:53
  • $\begingroup$ @Amin : Do you have a further response to my answer and comment? $\endgroup$
    – Iosif Pinelis
    Commented Jan 31, 2023 at 21:57
  • $\begingroup$ Based on the search, I found that $tr(A^{-1})tr(A)\le n^2 \kappa$ where $\kappa$ is the condition number of matrix $A$. In the following post, I defined matrix $V$ and I wanna find an upper bound on trace of inverse of $A$ that depends on $t$. My guess is that this upper bound is a decreasing function of $t$ (I guess that $t$ should appear in the denominator, not in the nominator). I should mention that vectors $\mathbf{x}_s$ are not chosen by adversaries. math.stackexchange.com/questions/4626468/… $\endgroup$
    – Amin
    Commented Feb 1, 2023 at 2:27
  • $\begingroup$ I generate samples and this behavior is observed $\endgroup$
    – Amin
    Commented Feb 1, 2023 at 2:28

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