# Operation in Hilbert space with matrices

Let $\{e_n\mid n \in \mathbb{N}\}$ be an orthonormal basis for the Hilbert space $H$ and define for each $T \in B(H)$ the doubly infinite matrix $A = \{\alpha_{n,m}\}$ by letting $\alpha_{n,m} = (T e_m\mid e_n)$.

1. Show that every row and every colomn in $A$ is square summable (i.e. belongs to $\ell^2$ ).
2. Use this to prove that the matrix product $AB = C$, where $C = \{\gamma_{n,m}\}$, where $\gamma_{n,m} = \sum_k \alpha_{n,k} \beta_{k,m}$ is well defined when matrices $A$ and $B$ correspond to operator $T$ and $S$ in $B(H)$.
3. Show that the matrix $C$ is corresponding to the operator $TS$.
4. Also find the matrix corresponding to $T + S$ and $T^*$.
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Where are you stuck? –  Davide Giraudo Nov 19 '12 at 18:13
@DavideGiraudo sorry to saying that but i am struggling even from the starting of this solution –  math Nov 19 '12 at 19:43
Don't be sorry, we are here to give you help. Do you know Bessel's inequality? –  Davide Giraudo Nov 19 '12 at 19:53
@DavideGiraudo yes i know, –  math Nov 19 '12 at 21:51

1. Bessel's equality says that for a Hilbert basis $\{e_n\}$ and $v\in H$, $\lVert v\rVert^2=\sum_{n=1}^{+\infty}|\langle v,e_n|^2$. Here, $$\sum_{n=1}^{+\infty}|\alpha_{n,m}|^2=\sum_{n=1}^{+\infty}|\langle Te_n,e_m\rangle|^2=\sum_{n=1}^{+\infty}|\langle e_n,T^*e_m\rangle|^2=\lVert T^*e_m\rVert^2,$$ so the sum of the column is finite. We do the same for the rows.
2. Well-definiteness is a consequence of $|\alpha_{n,k}\alpha_{k,m}|\leqslant \frac{|\alpha_{n,k}|^2+\alpha_{k,m}|^2}2$ and the first question.
3. Check $C\cdot e_n$ is what we want.