Suppose $\mathcal{H}$ is a separable Hilbert space over $\mathbb{C}$ (countable dimensions) with inner product $\langle,\rangle$. Let $A$ be a bounded linear operator on $\mathcal{H}$, i.e, in $B(\mathcal{H}$). Suppose further that $A$ is not a multiple of the identity operator. Then is it true that there exist two elements of $\mathcal{H}$, call them $v_1$,$v_2$, of norm 1, such that $\langle v_1 , A v_1 \rangle \neq \langle v_2, A v_2 \rangle$? This is true in finite dimensions (I think).
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$\begingroup$ For finite dimensional spaces, this is false. For instance, consider rotations in $R^2$. $\endgroup$– MalteCommented Jun 10, 2012 at 14:38
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$\begingroup$ I think he/she has a complex Hilbert space in mind. $\endgroup$– Michael RenardyCommented Jun 10, 2012 at 14:58
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$\begingroup$ yes, I mean complex Hilbert space, thanks for pointing out. Over ℝ, skew symmetric matrices are a counterexample in finite dimensions. $\endgroup$– magya_bloomCommented Jun 10, 2012 at 15:57
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2 Answers
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The answer is yes, this is true (assuming that the Hilbert space is complex).
If $\langle \xi,A\xi \rangle = \sigma$ for some $\sigma \in \mathbb C$ and all $\xi$, then $B:=A - \bar \sigma 1_H$ has the property that $\langle \xi,B\xi \rangle =0$ for all $\xi \in H$. We need to show $B=0$. Let $\xi \in H$ be arbitrary and consider the vector $\lambda \xi + \mu B\xi$ for some $\lambda,\mu \in \mathbb C$.
We get: $$0=\langle \lambda \xi + \mu B \xi, \lambda B\xi + \mu B^2 \xi \rangle = \lambda \bar\mu \langle \xi,B^2 \xi \rangle + \mu \bar\lambda \|B \xi\|^2$$ for all complex $\lambda$ and $\mu$. Taking $\lambda = \mu = 1$, we see $\|B\xi\|^2 = - \langle \xi,B^2 \xi \rangle$. Taking $\lambda=1, \mu=i$, we get $\|B\xi\|^2 = \langle \xi,B^2 \xi \rangle$. This shows $B \xi =0$.
answered Jun 10, 2012 at 15:56
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1$\begingroup$ Or you can just use the polarization identity ... $\endgroup$ Commented Jun 10, 2012 at 17:13
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$\begingroup$ Nik, the polarization identity gives the scalar product in terms of the norm. What would be the argument? Anyhow, since the argument is elementary, there are surely many ways to see this. $\endgroup$ Commented Jun 10, 2012 at 17:29
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4$\begingroup$ Andreas, if $\omega$ is uniformly distributed in the unit circle, or in $\{1,i,-1,-i\}$, I like to call the formula $\langle\xi,B\eta\rangle = \mathbb E(\omega \langle\xi+\omega \eta,B(\xi+\omega \eta)\rangle)$ a polarization identity. $\endgroup$ Commented Jun 10, 2012 at 18:01
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Nothing new compared to Andreas's answer, just wanted to stress the polarization idea:
Notation: For $H$ a Hilbert space, and $A\in B(H)$ (bounded linear operator), write $q_A$ for the quadratic form $x\mapsto \langle Ax,x\rangle$.
Lemma ('polarization'): If $H$ is a complex Hilbert space, $q_A=q_B\Leftrightarrow A=B$.
Proof: We may assume $B=0$ [replacing $A$ by $A-B$]. If $\langle Ax,x\rangle=0$ for all $x$, then $0=\langle A( x+y),x+y\rangle$ implies $\langle Ax,y\rangle+\langle Ay,x\rangle=0$. But then [replace x by ix] also $\langle Ax,y\rangle-\langle Ay,x\rangle=0$.
Answer to question: Yes, and separability is not needed. Proof by contraposition:
If $\lambda:=q_A(x)=q_A(y)$ for all $x,y$ of norm 1, then $q_A(h)=\lambda\|h\|^2=q_{\lambda I}(h)$ for all $h\in H$. Hence $q_A=q_{\lambda I}$, and the lemma implies $A=\lambda I$.
