# Positive compact operator has unique square root.

Let H be a hilbert space and T be a compact positive operator so that by the spectral decomposition theorem, $T=\sum_{n=1}^{\infty}{\lambda}_{n}\langle x,e_{n}\rangle e_{n}$ where the $e_{n}$ are the eigenvectors of T and form an orthonormal basis. I have shown that T has a positive square root namely $S=\sum_{n=1}^{\infty}{\lambda}_{n}^{0.5}\langle x,e_{n}\rangle e_{n}$. I am trying to show S is unique.

My attempt: Suppose that $L$ is positive with $L^2=T$.Then since $L$ commutes with $T$ it is invariant on it's eigenspaces. So given $e_{n}$ there is an $m$ such that $e_{m}$ is associated to the same eigenvalue, ${\lambda}$ as $e_{n}$ and $L(e_{n})={\alpha}e_{m}$ for some ${\alpha}$. Note that since each eigenvalue may be associated to more than one eigenvector, we cannot assume $n=m$. From there I'm stuck

Thanks

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Try $\langle S-L, e_n \rangle$. –  Eric Towers Mar 10 '14 at 9:34
@eric Towers. Do you mean $<(T-L)e_{n},{e_n}>$? Not really sure what that gives us. I don't know what L(e_n) is so no point expanding. It also gives us T-L is self adjoint but we already knew that –  user108605 Mar 10 '14 at 9:45
Nope. I'm mean "$S-L$". When given two things that must be the same, show that their difference is zero. In a "space" (with a basis) show that their projections onto the basis are zero in each component. –  Eric Towers Mar 10 '14 at 9:48
How do I show the inner product is zero? All I know is $S(e_{n})={\lambda}^{0.5}e_{n}$ and $S^2=L^2$ @Eric Towers –  user108605 Mar 10 '14 at 10:05
Well, $S^2 - L^2 = 0$ should give a hint. –  Eric Towers Mar 10 '14 at 10:07

## 1 Answer

Let $T$ and $S$ be as stated, and let $\{ \lambda_{j}\}$ be the distinct eigenvalues of $T$. Let $P_{j}$ be the orthogonal projection of $X$ on $\mathcal{N}(T-\lambda_{j}I)$. Then $Tx=\sum_{j=1}^{\infty}\lambda_{j}P_{j}x$ and $Sx=\sum_{j=1}^{\infty}\sqrt{\lambda_{j}}P_{j}x$.

Suppose $L\in\mathscr{L}(X)$ satisfies $0 \le L=L^{\star}$ and $L^{2}=T$. As you noted, $LT=TL$ must occur. Therefore, $$(T-\lambda_{j}I)LP_{j}=L(T-\lambda_{j}I)P_{j}=0,$$ which implies that $\mathcal{R}(LP_{j})\subseteq \mathcal{N}(T-\lambda_{j}I)=\mathcal{R}(P_{j})$ and, hence, $P_{j}LP_{j}=LP_{j}$. Because $L=L^{\star}$, then $LP_{j}=P_{j}L$. By the definition of $S$, it follows that $SL=LS$.

So, the problem reduces to the general problem of showing that two non-negative commuting operators $S$ and $L$ in $\mathscr{L}(X)$ are equal iff their squares are equal. So suppose that $S^{2}=T^{2}$. Because $S$ and $T$ commute, $$(S-L)(S+L)=S^{2}-L^{2}=0.$$ To argue that $S=L$, it is enough to show that $S=L$ on $\mathcal{N}(S+L)$ and on $\mathcal{R}(S+L)$ because $$X = \mathcal{N}(S+L)\oplus \mathcal{R}(S+L)^{c}.$$ (Here 'c' denotes topological closure.) Clearly $S-L=0$ on $\mathcal{R}(S+L)$ and, thus, also on $\mathcal{R}(S+L)^{c}$. Suppose that $x \in \mathcal{N}(S+L)$. For any $A=A^{\star} \ge 0$, one has $(Ax,x)=0$ iff $x=0$. Assuming $(S+T)x=0$ combines with $0 \le S \le S+T$ and $0 \le T \le S+T$ to imply that $Sx=0$ and $Tx=0$, which gives $S=L=0$ on $\mathcal{N}(S+T)$. So $S=T$.

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