Characterization of scalar operators on Hilbert spaces.

Question

Let $H$ be a complex Hilbert space. Let$A \in B(H)$.

I need to prove the following implication: If $A$ is not a scalar operator, then there exists a nonzero idempotent $P$ such that $AP$ is also a nonzero idempotent.

Answer 1

If $A$ is not a multiple of the identity there exists $v\in \mathcal{H}$ such that $Av$ and $v$ are linearly independent.

Lemma If $u$ and $v$ are linearly independent then there exists $w$ such that $$\langle w,u\rangle =\langle w,v\rangle =1,\qquad (*)$$ Proof Let $w=au+bv,$ $a,b\in \mathbb{C}.$ Then $(*)$ is equivalent to \begin{eqnarray*} a\|u\|^2+b\langle v,u\rangle &=&1\\ a\langle u,v\rangle+b\|v\|^2 &=&1 \end{eqnarray*} The determinant of the system of two equations is equal $D:=\|u\|^2\|v\|^2-|\langle u,v\rangle|^2.$ Hence it is positive, as the vectors $u$ and $v$ are linearly independent. Thus $(*)$ admits a solution. $\blacksquare$

By Lemma there is $w$ such that $$\langle v,w\rangle =1=\langle Av,w\rangle =1\qquad (**)$$ Let $Px=\langle x,w\rangle v.$ Then $APx=\langle x,w\rangle Av.$ Due to $(**)$ the operators $P$ and $AP$ are nonzero one-dimensional idempotents.