# Finite rank operators on Hilbert spaces

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.

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 $$\lambda_j$$.