Show that if $1<p< \infty$ and $T : l^p \rightarrow l^p$ is not compact, then there is a complemented subspace $E$ of $l^p$ such that $T$ is an isomorphism of $E$ onto a complemented subspace $T(E)$. Deduce that the Banach algebra $\mathcal{L}(l_p)$ contains exactly one proper closed two-sided ideal(the ideal of compact operators).
I am stuck with this problem. I am trying to use the theorem that if $T$ is compact then it's strictly singular, but that is true for only operators from $c_0$. I don't know how to use that. Any help is appreciated.
Here is my try. Suppose not, i.e., $\forall$ complemented subspace $E$ of $l_p$ $T|_E: E \rightarrow T(E)$ is NOT an isomorphism.
Let $Y$ be an infinite dimensional closed subspace of $l^p$.
Then $Y$ contains a subspace $E$ such that $E$ is isomorphic to $l^p$ and $E$ is complemented in $l^p$. So $T|_E : E \rightarrow T(E)$ is not an isomorphism.
That means $T|_Y:Y \rightarrow T(Y)$ is NOT an isomorphism. Which implies $T$ is strictly singular which implies $T$ is compact and which is a contradiction!
Is my proof correct?