If $T|_{[x]^{\bot}}=0$, then $T=u\otimes x$ for some $u\in H$?

Question

Suppose $H$ is a Hilbert space and $T\in B(H)$. And suppose $x\in H$ is such that $T|_{[x]^{\bot}}=0$ (Where $[x]$ denotes the subspace spanned by $\{x\}$). Is it true that $T=u\otimes x$ for some $u\in H$?

I can decompose $H$ as $H={[x]}\oplus [x]^{\bot}$ and then prove that the rank of $T$ is at most one. Then $T$ is of the form $h_1\otimes h_2$ for some $h_1,h_2\in H$. But how can I show that $h_2$ could be $x$?

Answer 1

If $T\equiv 0$, then yes, $T=0\otimes x$. So we assume $T$ is not the zero operator.

EDIT: A comment on why the rank is at most $1$. Let $P:H\to [x]$ be the orthogonal projection onto $[x]$ and let $Q=I_H-P$. Then $T=TI_H=T(P+Q)=TP+TQ=TP$. This is because the range of $Q$ is $[x]^\perp$, which is in the kernel of $T$, so $TQ=0$. Therefore $T=TP$, and $\text{rank}(T)\leqslant \text{rank}(P)\leqslant 1$.

As you said, we can write $T=h_1\otimes h_2$. Since $T$ is not the zero operator, $h_1,h_2\neq 0$.

Write $h_2=ax+u$ with $ax\in [x]$ and $u\in [x]^\perp$. We'll show that $u=0$. Note that since $u\in [x]^\perp$, $Tu=0$, and \begin{align*} 0 & = \langle 0,h_1\rangle = \langle Tu,h_1\rangle = \Bigl\langle \langle u,h_2\rangle h_1,h_1\Bigr\rangle = \Bigl\langle \langle u,ax\rangle h_1 + \langle u,u\rangle h_1,h_1\Bigr\rangle \\ & = \Bigl\langle 0+\langle u,u\rangle h_1,h_1\Bigr\rangle = \|u\|^2\|h_1\|^2.\end{align*} Since $h_1\neq 0$, $u=0$. Thus $h_2=ax$ for some scalar $a$. Since $h_2\neq 0$, $a\neq 0$, and $T=h_1\otimes ax = \overline{a}h_1\otimes x$.

EDIT $2$: We can skip the consideration of the rank of $T$ and the fact that $T=h_1\otimes h_2$ and go straight for the conclusion. If $x=0$, $[x]^\perp = H$ and $T\equiv 0$. Omitting that trivial case, fix $y=ax+u$, $ax\in [x]$, $u\in [x]^\perp$, and note that $a=\langle y,x\rangle/\|x\|^2$. Then $$Ty=T(ax+u)=Tax+0=T\langle y,x\rangle x/\|x\|^2 = \langle y,x\rangle \frac{Tx}{\|x\|^2}= \Bigl(\frac{Tx}{\|x\|^2}\otimes x\Bigr)(y).$$ This immediately yields that $T=\frac{Tx}{\|x\|^2}\otimes x$. Although I think $\frac{Tx}{\|x\|}\otimes \frac{x}{\|x\|}$ might be a more intuitive way of writing this.