The question is an extention to the answered question [prove spectral equivalence bounds for fractional power of matrices](https://mathoverflow.net/questions/425375/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?



prove spectral equivalence bounds for inverse fractional power of matrices
By [Remark 1 after Theorem 4.1](https://link.springer.com/book/10.1007/978-3-030-22422-6), 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 $0\le\alpha\le1$.


