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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 Reighleigh 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 $D^{-1}A.$

Now my question is: For $0 < \alpha \le 1,$ does $$ (c^-)^\alpha x^\top D^\alpha x \le x^\top A^\alpha x \le (c^+)^\alpha x^\top D^\alpha x $$ 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.

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$\newcommand\C{\mathbb C}\newcommand\R{\mathbb R}\newcommand\al{\alpha}$Yes, this follows by Loewner's theorem on monotone matrix functions (see e.g. Theorem 1.6), which in particular implies the following:

Let $M_n$ denote the set of all analytic functions $f\colon\C\setminus(-\infty,0]\to\C$ such that $f((0,\infty))\subseteq\R$ and $$A\le B\implies f(A)\le f(B)$$ for all $n\times n$ positive-definite matrices $A$ and $B$, where $A\le B$ means that $B-A$ is positive semidefinite. Then $f\in M_n$ for all natural $n$ if $$\Im z>0\implies \Im f(z)>0.$$

The above conditions on $f$ hold if $f(z)=z^\alpha$ for $\alpha\in(0,1]$ and all $z\in\C\setminus(-\infty,0]$.

So, your desired result immediately follows.

Even more immediately, your desired result follows from Theorem 4.1, which in turn follows from the identity $$x^\al=\frac{\sin\pi\al}\pi\int_0^\infty w^{\al-1}x(x+w)^{-1}\,dw$$ for real $x>0$, since in this identity $x$ can be replaced by any positive-definite matrix $A$ (see formula (4.5)), and the monotonicity of $A(A+w)^{-1}=I-w(A+w)^{-1}$ in $A$ for $w>0$ is easy to check (say, by differentiation).

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  • $\begingroup$ Thank you - again - for the very profound answer. I have never heard of this theorem before but obviously, it was exactly what I was looking for! $\endgroup$
    – Luna947
    Jun 24, 2022 at 12:32
  • $\begingroup$ You are welcome. Yes, Loewner's theorem is a great one. $\endgroup$ Jun 24, 2022 at 12:35

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