Uniqueness of square root for positive compact self-adjoint operator over Hilbert space.

Question

Let $T=\sum_{n=1}^{\infty}{\lambda}_{n}\langle x,e_{n}\rangle e_{n}$, be the compact, positive self-adjoint operator. My professor showed that the existence of the $S=\sum_{n=1}^{\infty}{\lambda}_{n}^{0.5}\langle x,e_{n}\rangle e_{n}$ which is compact, self-adjoint and positive.

However, I do not understand his explanation behind the uniqueness. Could someone please explain it to me:

Suppose $S'$ satisfies the conditions of the square root above i.e. $(S')^2=T$.

Then $S'$ commutes with T. (Clear to me).

Also $S'$ preserves the eigenspaces of T i.e if $T(e_n)=\lambda_ne_n$ then $T(S'(e_n))=\lambda_nS'(e_n)$ and $(S')^2=T=\lambda_n*Id$ on the eigenspace of $\lambda_n$. (Both Clear to me.)

Also the restriction of $S'$ on the eigenspace of $\lambda_n$ has an eigenvalue of $\sqrt \lambda_n$(This he proved using the fact that the restriction satisfies $(S')^2=T=\lambda_n*Id$ and $S'$ is positive)(This was ok).

However, he concludes by saying that the restriction of $S'$ on the eigenspace of $\lambda_n$ is $\sqrt \lambda_n*Id$. This I am not clear he just showed that $\sqrt \lambda_n$ is an eigenvalue. But how does he know that it applies to the whole space?

Edit: Since the restriction of $S'$ on the eigenspace of $\lambda_n$ has an eigenvalue of $\sqrt \lambda_n$ this implies that there is an $x$ in the eigenspace of $\lambda_n$ s.t $S'x=\sqrt \lambda_nx$. But that is only for that specific x. How do we know that it holds for all the other elements in the eigenspace?

Answer 1

If $x$ belongs to the eigen space of $\lambda_n$ then $S'x=\sqrt {\lambda_n}x$. The equation obviously holds if we take linear combinations. So $S'x=\sqrt {\lambda_n}x$ holds for every $x$ in the eigen space corresponding to $\lambda_n$. This means $S'=\sqrt {\lambda_n}I$ on this space.

Answer 2

If $T=S^2$ where $S :\mathcal{D}(T)\subseteq\mathcal{D}(S)\rightarrow\mathcal{D}(S)$, then the following makes sense and holds on all of $\mathcal{D}(T)$: $$ (S-\lambda I)(S+\lambda I)=T-\lambda^2I. $$ Therefore, if $Tx=\lambda^2x$ and $\lambda > 0$, $$ x = \frac{1}{2\lambda}(S+\lambda I)x-\frac{1}{2\lambda}(S-\lambda I)x $$ and $$ \frac{1}{2\lambda}(S+\lambda I)x\in\mathcal{N}(S-\lambda I) \\ \frac{1}{2\lambda}(S-\lambda I)x\in\mathcal{N}(S+\lambda I). $$ In order for $S$ to be positive, it is necessary that $\mathcal{N}(S+\lambda I)=\{0\}$, which forces $(S-\lambda I)x=0$. So $Tx=\lambda^2 x$ for some $x\ne 0$ and $\lambda >0$ forces $Sx=\lambda x$.