# Show compactness of an operator and calculate its SVD

Consider

$$T\colon\ell^2\to\ell^2, (s_1,s_2,\ldots)\mapsto (s_2,s_3,\ldots)$$

$$S\colon\ell^2\to\ell^2, (s_n)\mapsto (s_n/n)$$

$$R:=TS.$$

1) Show that $R$ is a compact operator.

2) Calculate the Singular Value Decomposition of $R$.

EDIT: My result concerning 2.)

The singular value decomposition $$\left\{(\sigma_j),(u_j),(v_j)\right\}$$ of $R$ is given by $$\sigma_j=\frac{1}{j+1},~~~~~v_j=e_j, j\geq 1,~~~~~u_j=e_j, j\geq 2,$$ where $\left\{u_j\right\}$ is an Orthonormalsystem of $\ell^2$ and $\left\{v_j\right\}$ is an Orthonormalsystem of $\ell^2$.

Is that right? Is that the SVD of $R$?

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No, this isn't correct.

In your fourth paragraph, you chose an arbitrary sequence $(s_n)$ from $\ell_2$. It need not be bounded (limited). Your argument just shows that each $s_n$ is in $\ell_2$. The constant $C$ you found depends of the particular $s_n$ that you are looking at. This error is easily remedied: you simply assume at the start that $(x_n)$ is a bounded sequence in $\ell_2$.

A more egregious error occurs later. The vector $u$ that you constructed is not a subsequence of $(R(s_n))$; it's just an element of $\ell_2$. So, it seems you tried to identify a candidate for a subsequential limit point; but, I don't think you can show that some subsequence of $(R(s_n))$ converges $u$. You need to be more careful with your diagonalization.

But, if you want to use a diagonalization argument, you can do the following:

Assume $(s_n)$ is a bounded sequence in $\ell_2$. If we can show that $(R(s_n))$ has a convergent subsequence, it will follow that $R$ is compact.

Towards this end, set $y_n=R(s_n)$. Then "diagonalize" $(y_n)$ as follows:

Choose a subsequence $(y^1_n)$ of $(y_n)$ so that its first coordinates form a convergent sequence in $\Bbb R$: $\lim\limits_{n\rightarrow\infty} y^1_n(1)=z_1$ for some number $z_1$.

Then choose a subsequence $(y^2_n)$ of $(y^1_n)$ so that its second coordinates form a convergent sequence in $\Bbb R$: $\lim\limits_{n\rightarrow\infty} y^2_n(2)=z_2$ for some number $z_2$.

Continue...

Now set $x_n=y_n^n$. Then $(x_n)$ is a subsequence of $(y_n)$ such that for each $i$ we have $\lim\limits_{n\rightarrow\infty} x_n(i)=z_i$.

Now show that $z\in \ell_2$ and that $\Vert x_n-z\Vert\rightarrow 0$. (Alternatively, you could just ignore $z$ and show that $(x_n)$ is Cauchy in $\ell_2$.)

I'll remark that the argument can be simplified a tad by noting that, since $T$ is bounded and since the composition of a bounded operator with a compact operator is compact, it suffices to show that $S$ is compact.

Showing that $S$ is compact can be done without diagonalizing explicitly by, for instance, showing that $S(B(\ell_1))$ is totally bounded. Or, by showing that $S$ is the norm limit of finite rank operators (see, this post, for example).

This post will also prove helpful.

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Thanks. And how can I calculate the SVD? –  math12 Feb 2 '13 at 19:48
Finding the adjoint operator $R^*$ and and then finding the Eigenvalues of $RR^*$? (resp. the Eigensequences?) –  math12 Feb 2 '13 at 20:14
@math12 You're welcome. I'm not familiar with SVD; so I'm afraid I can't help much with that. But, I will think about it. –  David Mitra Feb 2 '13 at 20:19
Is the adjoint operator of $R$ given by $R'\colon \ell^2\to \ell^2, (x_1,x_2,...)\mapsto \left(0,\frac{x_1}{2},\frac{x_2}{3},\frac{x_3}{4},...\right)$? –  math12 Feb 3 '13 at 11:21
@math12 That looks correct. You should verify that $(Rx,y)=(x,R^*y)$ for all $x,y$. –  David Mitra Feb 3 '13 at 12:35