Property of the selfadjoint operators with eigenvaluesin Hilbert spaces

Question

I have to prove this exercise:

Let $H$ be an Hilbert space, $T\in\mathcal L(H)$ a self-adjoint operator and $\lambda$ an eigenvalue of $T$.

1. The operator $T-\lambda I$ is self-adjoint, with $I$ being the identity operator.
2. $Ker(T-\lambda I)=Ker(T-\lambda I)^2$
3. Deduc from 2. that $Ker(T-\lambda I)=\bigcup_{m=1}^\infty Ker(T-\lambda I)^m$

My atempt:

I have proved 1. by deffinition of self-adjoint operator and the fact that if $\lambda$ is an eigenvalue of a self-adjoint operator then $\lambda$ is real. I also see that 3. is a trivial conclusion of 2..

For 2. I can see that the content '$\subset$' is trivial, but I have troubles with the other one. I see that if $x\in Ker(T-\lambda I)^2$, then $Tx-\lambda x$ is an eigenvector of $T$ corresponding to $\lambda$ eigenvalue. But I can't go on from here.

Answer 1

If $S$ is any self-adjoint operator and $S^{2}x=0$ then $ \langle S^{2}x, x \rangle=0$ implies $ \langle Sx, Sx \rangle=0$ which means $\|Sx\|^{2}=0$ so $Sx=0$. Take $S=T-\lambda I$.