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The question is an extention to the answered question prove spectral equivalence bounds for fractional power of matrices.

Let $A, D \in \mathbb{R}^{n \times n}$ be two symmetric,positive definite and tri-diagonal matrices for that we know that they are spectrally equivalent, thus ist holds $$ c^- x^\top D x \le x^\top A x \le c^+ x^\top D x $$ for any $x \in \mathbb{R}^n$, where $c^+, c^- > 0.$ The matrices $A$ and $D$ can be diagonalized, that is $$ A = V\Lambda_A V^\top, \quad D = W\Lambda_D W^\top $$ where $V$ and $W$ contain the eigenvectors of $A$ and $D$, and $\Lambda_A$ and $\Lambda_D$ are diagonal matrices containing the respective eigenvalues. Based on the Reileigh quotient, it should follow that $$ cond(D^{-1}A) \le \frac{c^+}{c^-},$$ thus $c^+$ and $c^-$ upper and lower bounds for the range of the eigenvalues of $D^{-1}A.$

In the question I linked above, I got the answer that due to Loewner's theorem, for $0 < \alpha \le 1,$ $$ (c^-)^\alpha x^\top D^\alpha x \le x^\top A^\alpha x \le (c^+)^\alpha x^\top D^\alpha x $$ does hold. Here, $A^\alpha := V\Lambda_A^\alpha V^\top,$ and $D^\alpha := W\Lambda_D^\alpha W^\top,$ where $\Lambda_A^\alpha, \Lambda_D^\alpha$ can be computed by taking the power $\alpha$ of each diagonal entry.

Now my question is:

Is it possible to deduce the spectral bound estimates for the inverse of the matrices, that is for $A^{-1}$ and $D^{-1}$ as well as $A^{-\alpha}$ and $D^{-\alpha}$ in the same manner ? I expect something like, e.g., $$\frac{1}{c^+} x^\top D^{-1} x \le x^\top A^{-1} x \le \frac{1}{c^-} x^\top D^{-1} x $$ and $$\frac{1}{(c^+)^\alpha} x^\top D^{-\alpha} x \le x^\top A^{-\alpha} x \le \frac{1}{(c^-)^\alpha} x^\top D^{-\alpha} x $$ that holds for any $x \in \mathbb{R}^n$.

And second: Is it possible to deduce the condition numbers for $(D^{-1} A)$ and $(D^{-\alpha} A^\alpha)$ from that?

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By Remark 1 after Theorem 4.1, the matrix expression $-A^{-\alpha}$ is Loewner-nondecreasing in positive definite matrix $A$ if and only $0\le\alpha\le1$.

So, the inequalities $$\frac{1}{(c^+)^\alpha} x^\top D^{-\alpha} x \le x^\top A^{-\alpha} x \le \frac{1}{(c^-)^\alpha} x^\top D^{-\alpha} x $$ will hold in general if and only $0\le\alpha\le1$.

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  • $\begingroup$ Thank you for the hint. I have seen that remark, but in proposition 2.2 (to which they refer in that remark) they state that in that case, the eigenvalues of both A and D have to be in (-1,0). In our case, due to positive definiteness of A and D, this does obviously not hold. So the statement does not hold? $\endgroup$
    – Luna947
    Jun 27 at 0:09
  • $\begingroup$ @Luna947 : You should apply Proposition 2.2 with $A,B$ replaced by $-A,-B$, or by $-A^{-1},-B^{-1}$. In other words, if $0<A\le B$ (in the Loewner sense), then $A^{-1}\ge B^{-1}$ -- which is what one should use here. $\endgroup$ Jun 27 at 1:11
  • $\begingroup$ ok, so I still have one question: in our case, $A$ and $B$ have eigenvalues in (0,\infty ). so the eigenvalues of $-A$ and $-B$ are in (-\infty ,0) and for $-A^{-1}$ and $-B^{-1}$ they are in in (-\infty ,0) and (-\infty ,0). How can I now deduce the statement using Proposition 2.2 since there it is neccessary to have the eigenvalues in the range $(-1,0)$ ? $\endgroup$
    – Luna947
    Jun 27 at 8:55
  • $\begingroup$ And let me come up with another question - i think that should be the last one! Is it possible to make a statement concerning the spectral equivalence estimates of the sum, e.g., something like $(A+A^{-1}) \le \max\{c^+, \frac{1}{c^+}\} (D + D^{-1})$ based on the bounds that we deduced so far?? $\endgroup$
    – Luna947
    Jun 27 at 10:41
  • $\begingroup$ @Luna947 : no, this is impossible. $\endgroup$ Jun 27 at 14:17

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