Properties of compact operator in $l_2$ commuting with compacts
Can you please explain why from $ ||Ke_i||=||Ke_j||  \forall i,j$
follows that K compact only if K=0?
 A: Let $v$ be any vector in $\ell^2$. 
Then $\langle v,\,Ke_i\rangle=\langle K^*v,\,e_i\rangle \rightarrow 0$ as $i$ goes to infinity, because $K^*v$ is a vector $\ell^2$ as well.
Thus $Ke_i$ converges weakly to zero. Now we know that this sequence is precompact in $\ell^2$ with the strong topology, because $K$ is compact. Let $u$ be any limit point of the sequence: by the above, $u$ is orthogonal to any $v$, thus $u=0$. 
Since $\|Ke_i\|$ is constant, it is equal to $\|u\|=0$. Therefore $Ke_i=0$ for all $i$ thus $K=0$. 
A: The first answer is more general; this one keeps the context of the referred question.
Suppose $Ke_1=(a_1,a_2,\ldots)\in\ell_2$. Then $SKe_2=KSe_2=Ke_1$, so $Ke_2=(b,a_1,a_2,\ldots)$. But since $\|Ke_2\|=\|Ke_1\|$, it follows that $b=0$. By induction, $Ke_n=(0,\ldots,0,a_1,a_2,\ldots)$.
If $K$ is compact then the sequence $Ke_n$ has a convergent subsequence. But the only sequence that $Ke_{n_r}$ can possibly converge to is $0$, in which case $\|Ke_1\|=\|Ke_n\|\to0$ gives $a_i=0$ for all $i$, that is $Ke_n=0$ for all $n$ and $K=0$.
