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Let $H$ be a Hilbert space.

Question 1: Are all rank one operators from $H$ to $H$ is of the form $$T:H\rightarrow H, x \mapsto \langle x,u\rangle v $$ For some $u,v \in H$.

Question 2: Suppose $I \subseteq L(H)$ is an ideal and contains all the rank one operators, how do we show it contains all the finite rank operators?

These two statements seem to be true, but I could not find any reference.


After some thought: Let us fix orthonormal basis $\{u_i\}$ of $H$. We have two observations:

  1. Operator $T^*$ exists. So $$ Tx = \sum \langle Tx , u_i \rangle u_i = \sum \langle x, T^*u_i \rangle u_i $$
  2. $$x \mapsto \langle x,v \rangle w$$ are rank one if $v \not=0, w \not= 0$.

  3. Combining the above two, $T$ is rank one if and only if it is of the form $x \mapsto \langle x,v \rangle w $.

  4. Any finite rank operator, must again be of the form $\sum_j \langle x, v_j \rangle w_j$ (finite sum). These are generated by the rank one operators.


I would be happy if anyone point some possible pitfalls / mistake I made in my proof.

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I don't really see how you combine your 1 and 2 to get that $T$ is of the desired form when it is rank-one, so I cannot comment on that. I also don't see how you reason on 4.

If $T$ is rank-one, then there exists a fixed $y\in H$ with $\|y\|=1$ such that $Tx=\lambda(x)\,y$ for all $x$. From $y\ne0$ you get that the number $\lambda(x)$ is unique for each $x$. Now use the linearity of $T$ to deduce that $\lambda $ is linear. Also, $$ |\lambda(x)|=\|\lambda(x)y\|=\|Tx\|\leq\|T\|\,\|x\|. $$ So $\lambda$ is a bounded linear functional. By Riesz's Representation Theorem, there exists $z\in H$ with $\lambda(x)=\langle x,z\rangle$. Thus $$ Tx=\langle x,z\rangle y. $$

When $T$ is finite-rank, you can repeat the above but, instead of a single $y$, you will now have an orthonormal basis $y_1,\ldots,y_n$ and bounded linear functionals $\lambda_j$.

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