# Understanding rank $1$ operators on Hilbert Space

If the range of an operator $$T$$ is one-dimensional, then it is said to have $$\newcommand{\rank}{\operatorname{rank}}\rank 1$$ as stated in N.Young's book An Introduction to Hilbert Space, pg.84. Also, if $$T$$ is a bounded operator of $$\rank 1$$ on a Hilbert Space $$H$$, then $$Tx = \langle x, \phi \rangle \psi$$ for all $$x\in H$$ where $$\psi$$ is a non-zero vector in range of $$T$$ and $$\phi$$ is a fixed unique element of $$H$$.

So, $$\psi = Ty$$ for some $$y\in H$$, but then $$Tx= \langle x, \phi \rangle Ty$$. And this goes on forever, $$Tx= \langle x, \phi \rangle \langle y, \phi \rangle Tz$$... So, $$T$$ becomes an infinite product. What do I miss? What is the exact definition of $$\rank 1$$ operator? Thanks in advance.

• Let $\{e_n\}$ be an orthonornal basis for $H$. If $T$ has rank $1$, then the set $\{Tf:f\in H\}$ is spanned by $e_n$ for some $n$, that is, for any $f\in H$, $Tf = ce_n$ where $c\in\mathbb C$. Dec 5, 2019 at 3:25
• As I commented to @daw, I think $Tx=c_xe_n$ for $x\in H$. So, $Tx=e_n$ implies $c_x=1$. I still can not comprehend the rule of $T$. What does it do the elements of $H$? Is $T$ in dual space of $H$? Dec 6, 2019 at 3:36
• $T$ maps the elements of $H$ into the span of $e_n$ for one particular $n$. In other words, the image of $H$ under $T$ is $$T(H) = \{ce_n : c\in\mathbb C\}.$$ Dec 6, 2019 at 3:41

Let $$Tx=\langle x,\phi\rangle\psi$$. If $$y$$ is such that $$Ty=\psi$$ then $$\langle y,\phi\rangle=1$$. Then $$Tx = \langle x,\phi\rangle Ty = \langle x,\phi\rangle \langle y,\phi\rangle Ty = \dots$$ so all additional factors are $$1$$ and these extra points $$Tz$$ are all equal to $$Ty$$, $$Tz=Ty$$.
• Is $Ty$ a basis vector of $\mathbb{C}$? Here, we assume $T$ has range in $\mathbb{C}$ I think. Dec 5, 2019 at 12:32
• $Ty$ is a basis vector of the one-dimensional space $R(T)$.
• I think you mean $Ty = Tz$. Dec 5, 2019 at 13:37
• Combining your answer with @Math1000's, this $c$ in $ce_n$ changes with respect to $x$. A very interesting situation for me. Dec 5, 2019 at 13:38