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In this question, it uses the property \begin{align*} \max_{x\neq 0}\frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}= \max_{x\neq 0}\frac{(Ax,x)}{(x,x)} \end{align*} where $A,\sqrt{B}$ are invertible symmetric matrices and $x\in \mathbb{R}^n$.

I am thinking about how one would prove this. I provide a proof below using contradiction but language seems cumbersome to me. I am wondering whether a more concise proof is possible.

Proof

Suppose not. Then either \begin{align*} \max_{x\neq 0}\frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}< \max_{x\neq 0}\frac{(Ax,x)}{(x,x)} \end{align*} or \begin{align*} \max_{x\neq 0}\frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}> \max_{x\neq 0}\frac{(Ax,x)}{(x,x)} \end{align*} Let's just consider the first case as the other case would be similar.

Let $x_0=\arg \max_{x\neq 0}\frac{(Ax,x)}{(x,x)}$ and $x_1=\arg \max_{x\neq 0}\frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}$. Then, \begin{align*} \frac{(A\sqrt{B}x_1,\sqrt{B}x_1)}{(\sqrt{B}x_1,\sqrt{B}x_1}< \frac{(A\sqrt{B}\sqrt{B}^{-1}x_0,\sqrt{B}\sqrt{B}^{-1}x_0)}{(\sqrt{B}\sqrt{B}^{-1}x_0,\sqrt{B}\sqrt{B}^{-1}x_0}=\frac{(Ax_0,x_0)}{(x_0,x_0)} \end{align*} This means $x_1\neq \arg \max_{x\neq 0}\frac{(A\sqrt{B}x,\sqrt{B}x)}{(\sqrt{B}x,\sqrt{B}x)}$ which gives us a contradiction.

chuck
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  • Is $\sqrt{B}$ necessarily a matrix such that there is a matrix C s.t C^2 = B? – gist076923 Sep 25 '22 at 22:41
  • In the linkded question, yes. I am not sure whether that is needed in general. – chuck Sep 25 '22 at 22:47
  • Let $$f(x)={(Ax,x)\over (x,x)}$$ We have $x\neq 0$ iff $y=\sqrt{B}x\neq 0$ and $x=\sqrt{B}^{-1}y.$ Thus $$\max_{x\neq 0}f(x)= \max_{\sqrt{B}x\neq 0}f(\sqrt{B}x) =\max_{x\neq 0}f(\sqrt{B}x)$$ The matrix. $A$ does need to be invertible. – Ryszard Szwarc Sep 26 '22 at 00:47
  • $\frac{1}{|\sqrt{B}x|}\sqrt{B}x$ ranges over all unit vectors in the space. So, yes, the first equality in your post is an equality. – Disintegrating By Parts Sep 26 '22 at 09:12

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