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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.

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    $\begingroup$ 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 $c\in\mathbb C$. $\endgroup$
    – Math1000
    Dec 5 '19 at 3:25
  • $\begingroup$ 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 $H$? $\endgroup$
    – Tedebbur
    Dec 6 '19 at 3:36
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    $\begingroup$ $T$ 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 $$T(H) = \{ce_n : c\in\mathbb C\}. $$ $\endgroup$
    – Math1000
    Dec 6 '19 at 3:41
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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 $Ty$, $Tz=Ty$.

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  • $\begingroup$ Is $Ty$ a basis vector of $\mathbb{C}$? Here, we assume $T$ has range in $\mathbb{C}$ I think. $\endgroup$
    – Tedebbur
    Dec 5 '19 at 12:32
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    $\begingroup$ $Ty$ is a basis vector of the one-dimensional space $R(T)$. $\endgroup$
    – daw
    Dec 5 '19 at 12:40
  • $\begingroup$ I think you mean $Ty = Tz$. $\endgroup$
    – Tedebbur
    Dec 5 '19 at 13:37
  • $\begingroup$ Combining your answer with @Math1000's, this $c$ in $ce_n$ changes with respect to $x$. A very interesting situation for me. $\endgroup$
    – Tedebbur
    Dec 5 '19 at 13:38

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