# prove spectral equivalence bounds for inverse fractional power of matrices

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?

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 $$0\le\alpha\le1$$.
• @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. Jun 27 at 1:11
• 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 $(-1,0)$ ? Jun 27 at 8:55
• 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?? Jun 27 at 10:41