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suppose

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}$).$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> v_1 \rangle \neq < \langle v_2, A v_2>$. v_2 \rangle$? This is true in finite dimensions (I think).

2 added 18 characters in body

suppose $\mathcal{H}$ is a separable Hilbert space over $\mathbb{C}$ (countable dimensions) with inner product $<,>$. 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 $< v_1 , A v_1> \neq < v_2, A v_2>$. This is true in finite dimensions (I think).

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# non-Identity operator on a separable Hilbert space

suppose $\mathcal{H}$ is a separable Hilbert space (countable dimensions) with inner product $<,>$. 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 $< v_1 , A v_1> \neq < v_2, A v_2>$. This is true in finite dimensions (I think).