non-Identity operator on a separable Hilbert space - MathOverflow

non-Identity operator on a separable Hilbert space

2012-06-10T13:53:52Z

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).

Answer by Andreas Thom for non-Identity operator on a separable Hilbert space

2012-06-10T15:56:08Z

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 $B \xi =0$.

Answer by wildildildlife for non-Identity operator on a separable Hilbert space

2012-06-10T22:42:48Z

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 $A=\lambda I$.