# About the operator theory in Hilbert space

Let $H$ be a hilbert space. $L(H)$ be the set of linear operators on $H$.

Suppose that $S,T\in L(H)$ and $S\geq0$, $\|Sx\|=\|Tx\|$ for every $x\in H$.

Can I conclude that $S=\sqrt{T^*T}$?

-
Do I miss something obvious. Why it cannot be that $S=T$? –  Fabian Apr 5 '11 at 18:31
I modified it. Thanks. –  user8484 Apr 5 '11 at 18:34

Because $\|\sqrt{T^*T}x\|^2=\langle T^*Tx,x\rangle=\|Tx\|^2$ for all $x$, the question amounts to the following: If $S$ and $P$ are positive operators on $H$ such that $\|Sx\|=\|Px\|$ for all $x\in H$, then must $S=P$? The answer is yes.

For all $x$ and $y$ in $H$, comparing the expansions of $\|S(x+y)\|^2$ and $\|P(x+y)\|^2$ in terms of the inner product shows that $\Re\langle S^2x,y\rangle =\Re\langle P^2x,y\rangle$, and comparing the expansions of $\|S(x+iy)\|^2$ and $\|P(x+iy)\|^2$ in terms of the inner product shows that $\Im\langle S^2x,y\rangle=\Im\langle P^2x,y\rangle$. Therefore $\langle S^2x,y\rangle=\langle P^2x,y\rangle$ for all $x$ and $y$, which implies that $S^2=P^2$. By uniqueness of positive square roots, $S=P$.

-

Yes. For all $x \in H$, $\langle x, S^2 x \rangle = \|S x\|^2 = \|T x\|^2 = \langle x T^* T x \rangle$. By the polarization identity, $\langle x, S^2 y \rangle = \langle x, T^* T y \rangle$ for all $x$ and $y$, so that $S^2 = T^* T$. By the uniqueness of positive semidefinite square roots, $S = \sqrt{T^* T}$.

-
Yes: the condition $\|Sx\| = \|Tx\|$ squares and expands to $(S^*Sx|x) = (T^*Tx|x)$ for all $x$. It follows by Polarisation that $T^*T = S^*S = S^2$. Now positive operators have unique square-roots, so we conclude that $S = \sqrt{T^*T}$ as required.
Polarisation: for an operator $R$ we have that $$4(Rx|y) = \sum_{k=0}^3 i^k (R(x+i^ky)|x+i^ky).$$ So if $(Rx|x)=0$ for all $x$, this shows that also $(Rx|y)=0$, so $R=0$.
I think you're missing a factor of $1/4$ in you polarization identity? –  joriki Apr 5 '11 at 19:24