Polar Decomposition and Compact operator

Question

Let $H$ be an infinite dimensional separable Hilbert space and $\{e_n\}$ be a countable orthonormal basis for $H$.For a bounded sequence $\{a_n\}$ define $T(e_n)=a_ne_{n+1}$ and extend linearly to $\operatorname*{span}{\{e_n\}}$. Verify that $T$ extends to a bounded linear operator on $H$.Compute the adjoint of $T$ and polar decomposition of $T$. Find conditions on $\{a_n\}$ so that $T$ is a compact operator. I have verified that $T$ extends to $H$,but I am unable to do the other parts. Any help will be appreciated.

Answer 1

Let $T_{n}x = a_{n}(x,e_{n})e_{n+1}$. Then $$ (T_{n}x,y) = (a_{n}(x,e_{n})e_{n+1},y)=a_{n}(x,e_{n})(e_{n+1},y)=(x,\overline{a_{n}}(y,e_{n+1})e_{n}). $$ So $T_{n}^{\star}y=\overline{a_{n}}(y,e_{n+1})e_{n}$, and $T_{n}^{\star}T_{n}=|a_{n}|^{2}(x,e_{n})e_{n}$. The unique positive square root of $T_{n}^{\star}T_{n}$ is $|T_{n}|x = |a_{n}|(x,e_{n})e_{n}$. That should get your started.

Answer 2

Assuming that the inner products are linear with respect to the 2nd coordinate, let me introduce the bra-ket notation. Let $x \in H$, then $\left|x\right> \colon \mathbb{C} \rightarrow H$ if defined by $\left|x\right> \alpha = \alpha x$ for all $\alpha \in \mathbb{C}$, the operator $\left< x \right| \colon H \rightarrow \mathbb{C}$ is defined by $\left< x \right| y= \left<x, y \right>$ for all $y \in H$. Clearly, $\left<x\right|^* = \left|x\right>$.

Yout $T$ is given by the formula $T = \sum_{n \geq 0} a_n \left|e_{n+1}\right>\left<e_n\right|$, so we see that it is approximated by final rank operators, it is bounded if and only if $(a_n)$ is bounded it is compact if and only if $(a_n)$ is from $c_0$. The adjoint is given by $T^* = \sum_{n \geq 0} \overline{a_n} \left|e_{n}\right>\left<e_{n+1}\right|$. From here it is also easy to get the polar decomposition.