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The following is a fact in Murphy's C*-algebras and operator theory page 49:enter image description here

Suppose $u,v \in B(H)$, where $H$ is a Hilbert space, then $u=v$ if and only if $\langle u\xi,\xi\rangle = \langle v\xi,\xi\rangle$ for all $\xi \in H$.

Clearly if $u=v$ then $\langle u\xi,\xi\rangle = \langle v\xi,\xi\rangle$ for all $\xi\in H$. For converse, I should show $\langle u\xi,\eta\rangle=\langle v\xi,\eta\rangle$ for all $\xi,\eta\in H$, while I can not show it. I think about Polarisation identity but I can not use it. Please help me. Thanks in advance.

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    $\begingroup$ This is false as it stands. You need to require symmetry of your operators. $\endgroup$ – Giuseppe Negro Dec 20 '14 at 15:21
  • $\begingroup$ @GiuseppeNegro : It's a fact in a Murphy's book $\endgroup$ – niki Dec 20 '14 at 15:23
  • $\begingroup$ @MikeMiller : I think for the vector $\xi=(1+i,i)$, $\langle(u-v )\xi,\xi\rangle \neq 0$. So It's nota counterexample. $\endgroup$ – niki Dec 20 '14 at 15:35
  • $\begingroup$ @MikeMiller : Sorry, I can not understand your mean $\endgroup$ – niki Dec 20 '14 at 15:44
  • $\begingroup$ @MikeMiller : He did not prove it, I put the paragraph of the book above. $\endgroup$ – niki Dec 20 '14 at 15:48
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This is a corollary of the previous statement in the book: the polarization identity says that for a sesquilinear form $\sigma$, you can write $$4\sigma(x,y) = \sigma(x+y,x+y)+i\sigma(x+iy,x+iy) - \sigma(x-y,x-y) -i\sigma(x-iy,x-iy).$$ In particular, this shows that given two sesquilinear forms $\sigma, \sigma'$, if $\sigma(v,v) = \sigma'(v,v)$ for all $v$, the RHS is the same if you replace $\sigma$ and $\sigma'$; so $\sigma'(x,y) = \sigma(x,y)$. Thus it suffices to check if $\sigma = \sigma'$ on pairs $(v,v)$.

It's easy to check that the maps $\sigma(x,y) = \langle ux, y\rangle$ and $\sigma'(x,y) = \langle vx, y\rangle$ are sesquilinear forms. By hypothesis, you know that $\sigma(x,x) = \sigma'(x,x)$ for all $x$; so by the above discussion, $\sigma(x,y) = \sigma'(x,y)$ for all $x,y$. Rewriting this, we have $\langle (u-v)x, y \rangle = 0$ for all $x,y$; so picking $y = (u-v)x$, we see that $\|(u-v)x\|^2 = 0$, and thus that $(u-v)x = 0$. So $u=v$ as desired.

(This is really just a rephrasing of what Murphy wrote.)

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  • $\begingroup$ Oh, I made a mistake. However, Thanks for your attention. $\endgroup$ – niki Dec 20 '14 at 16:04
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WARNING. This is not a counterexample as it works only in real Hilbert spaces. See comments.


The statement is false if $u$ and $v$ are not assumed to be symmetric. Consider the Hilbert space $\mathbb{R}^2$. The operators $$ u\mathbf{x}=(-x_2,x_1)$$ and $$ v\mathbf{x}=(-2x_2,2x_1)$$ are such that $$ (u\mathbf{x}, \mathbf{x})=(v\mathbf{x}, \mathbf{x})=0,\qquad \forall \mathbf{x}$$ but clearly they differ.

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    $\begingroup$ It's true for complex spaces, and apparently only complex spaces are considered in the book. If you look at $u(1,i)$, you see that $(ux,x) \not\equiv 0$. $\endgroup$ – Daniel Fischer Dec 20 '14 at 16:10
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    $\begingroup$ The inner product on $\Bbb C^2$ is $\bf{x} \cdot \overline{\bf{y}}$. In particular, $\langle (u-v)x, x\rangle)$ needn't be zero; it doesn't kill eg $(1+i,i)$. (This is the OP's counterexample to my claim in the comments that this was a counterexample, which I later deleted.) $\endgroup$ – user98602 Dec 20 '14 at 16:10
  • $\begingroup$ @MikeMiller: Oh, I didn't ever notice this. Interesting. I'll leave this answer on with a warning sign, for reference. $\endgroup$ – Giuseppe Negro Dec 20 '14 at 16:42

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