Is $\langle Tu, u \rangle = 0$ true for *some* unit vector $u$ on any operator $T$ in a Hilbert space?

Question

I am currently studying the numerical range of an operator, $Num(T) = \{\langle Tu, u \rangle : u\in Dom(T), \|u\|=1\}$, and would like to know when $0 \in Num(T)$. This question is essentially the inverse of a question like In a complex vector space, $\langle Tx,x \rangle=0 \implies T = 0$ :

Given an arbitrary operator T on a Hilbert space, does there always exist some unit vector $u$ (i.e. with $\|u\| = 1)$ such that $\langle Tu, u \rangle = 0$? So far, I have unsuccessfully tried:

• showing that there is always some unit vector $u$ such that $Tu = 0$ (and thus $\langle Tu, u \rangle = \langle 0, u \rangle = 0$)
• showing that there is always some unit vector $u$ such that $T$ maps $u$ to a vector orthogonal to $u$.

Or alternatively, are there specific conditions where it is the case? Or is it never true (unless T = 0)?

Answer 1

An obvious counterexample is the identity operator: if $Tu=u$ for all $u$, then the numerical range is $\{1\}$.

It is easy to see that for $a,b\in\mathbb C$, you have $$\def\num{\operatorname{Num}}\num(aT+bI)=b+a\num(T).$$ As $\num(T)$ is always bounded when $T$ is bounded, this shows that for any bounded $T$ you can find $b$ such that $0\not\in\num(T+bI)$. And you can also find $b$ such that $0\in\num(T+bI)$.

Answer 2

does there always exist some unit vector u (i.e. with ∥u∥=1) such that ⟨Tu,u⟩=0?

I think this is trivially false, by counterexample: for an operator $T \in \mathbb{C}^{n\times n}$, with a positive spectrum, then in the diagonal basis of $T$, $\langle u, Tu\rangle = \sum_i \vert{u_i}\vert^2 T_{ii}$ (it's a sum of positive terms). For this to equal $0$, for all $i$, $u_i=0$ or $T_{ii}=0$. If $T_{ii} \neq 0$ for all $i$ (which you can easily construct), then obviously no $u \; : \; \Vert u \Vert = 1$ satisfying this exists.

Or alternatively, are there specific conditions where it is the case?

You could also construct an example for this case following the same line of reasoning: construct a (square, Hermitian) operator $T$ where two of its eigenvalues take values $\lambda$ and $-\lambda$, for some $\lambda > 0$. Let the corresponding eigenvectors be $w_1$, $w_2$. ($\langle w_1, w_2 \rangle = 0$, since they are eigenvectors.) Then $ u = \frac{1}{\sqrt{2}} (w_1 + w_2)$ is such that $\langle u, Tu \rangle = 0$, $\Vert u\Vert = 1$.