If $A$ is a positive linear operator on $H$ s.t. $A$'s eigenvectors form an ONB with eigenvalues $c_j$, are $\sqrt{c_j}$s eigenvalues of $\sqrt{A}$?

Question

Let $H$ be a Hilbert space over $\mathbb{C}$ and $A$ be a positive linear operator on $H$ (not necessarily bounded). Suppose that $A$'s eigenvectors $\{e_j\}$ form an ONB for $H$ and $\{c_j\}$s are the corresponding eigenvalues of $A$. We know from spectral theory that $A$'s square root $\sqrt{A}$ exists and is positive since $A$ is positive. But is it now true that $\{\sqrt{c_j}\}$s are eigenvalues of $\sqrt{A}$?

If I recall correctly, in the cast that $A$ is bounded we have that if $\sqrt{A}$ exists, it is self-adjoint. Therefore we know that for every $e_j$, $c_j = \left<Ae_j, e_j\right> = ||\sqrt{A}e_j||^2$ whence $\sqrt{c_j}\leq ||\sqrt{A}||$. But I don't really know how to turn this approach to a proof that $\sqrt{A}e_j = \sqrt{c_j}e_j$.

Answer 1

How about reasoning along the following lines. Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$, and let $$ w = \sqrt{A}v.$$ By looking at the matrix-representation of $\sqrt{A}|_{\operatorname{span}(v,w)}$ we easily find the eigenvectors $\sqrt{\lambda}v+w$ and $\sqrt{\lambda}-w$. It is easily check that $$\begin{align*} \sqrt{A}(\sqrt{\lambda}v+w) &= \sqrt{\lambda}\sqrt{A}v + Av \\ &= \sqrt{\lambda}w+\lambda v \\ &= \sqrt{\lambda}(\sqrt{\lambda}v + w) \end{align*}$$

Answer 2

I am including this comment as an answer and will accept it as soon as I can since it proves the claim. Inspiration for this proof is by @Testcase.

Take $w = \sqrt{c_j}e_j - \sqrt{A}e_j$. We aim to show that $||w|| = 0$. To that end,

$$\sqrt{A}w=\sqrt{c_j}\sqrt{A}e_j - Ae_j = \sqrt{c_j}(\sqrt{A}e_j - \sqrt{c_j}e_j) = -\sqrt{c_j}w$$

We assumed that $\sqrt{A}$ is positive. Therefore by $\left<w,w\right>\geq 0$, $\left<\sqrt{A}w,w\right>=-\sqrt{c_j}||w||^2\leq 0$. But this is only possible if $\sqrt{c_j}=0$ or $||w||=0$. Thus the claim follows.

Answer 3

We have $$A-\lambda I=(\sqrt{A}+\sqrt{\lambda}I)(\sqrt{A}-\sqrt{\lambda}I)$$ For $\lambda>0$ the first factor is an invertible operator. Therefore $(A-\sqrt{\lambda}I)v= 0$ iff $(A-\lambda I)v=0.$

Concerning $\lambda=0$ we have $\ker A=\ker A^{1/2}$ as $\langle Av,v\rangle =\|A^{1/2}v\|^2.$

The reasoning can be applied to $A^{1/n}$ by applying the formula for $b^n-a^n.$