non-Identity operator on a separable Hilbert space - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T00:09:32Z http://mathoverflow.net/feeds/question/99239 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99239/non-identity-operator-on-a-separable-hilbert-space non-Identity operator on a separable Hilbert space magya_bloom 2012-06-10T13:53:52Z 2012-06-10T22:42:48Z <p>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). </p> http://mathoverflow.net/questions/99239/non-identity-operator-on-a-separable-hilbert-space/99244#99244 Answer by Andreas Thom for non-Identity operator on a separable Hilbert space Andreas Thom 2012-06-10T15:56:08Z 2012-06-10T15:56:08Z <p>The answer is yes, this is true (assuming that the Hilbert space is complex).</p> <p>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 $\lambda,\mu \in \mathbb C$.</p> <p>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 $B \xi =0$.</p> http://mathoverflow.net/questions/99239/non-identity-operator-on-a-separable-hilbert-space/99263#99263 Answer by wildildildlife for non-Identity operator on a separable Hilbert space wildildildlife 2012-06-10T22:42:48Z 2012-06-10T22:42:48Z <p>Nothing new compared to Andreas's answer, just wanted to stress the polarization idea:</p> <p><strong>Notation</strong>: For $H$ a Hilbert space, and $A\in B(H)$ (bounded linear operator), write $q_A$ for the quadratic form $x\mapsto \langle Ax,x\rangle$. </p> <p><strong>Lemma</strong> ('polarization'): If $H$ is a complex Hilbert space, $q_A=q_B\Leftrightarrow A=B$.</p> <p><strong>Proof</strong>: 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 $\langle Ax,y\rangle-\langle Ay,x\rangle=0$.</p> <p><strong>Answer to question</strong>: Yes, and separability is not needed. Proof by contraposition:</p> <p>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 $A=\lambda I$.</p>