Polynomial and operator in Hilbert space

Question

I’m non able to show point 2 of this exercise:

Let $H$ be a Hilbert space and $T : H \to H $ a compact self-adjoint operator. Suppose there exists a polynomial $p: \mathbb{R} \to \mathbb{R}$ with only real zeros s.t. $p(T)=0$. Show

1. If $dim(H) = \infty $ then $0$ is an eigenvalue of $T$
2. If $p(s)>0$ when $s<0$, then $\langle Tx, x \rangle \ge 0$ for all $x \in H$

I’ve proved 1 but I have no idea of how to prove 2!

Answer 1

2. The polynomial $p$ maps the spectrum of $T$ to $0$ because $\{0\}=\sigma(p(T))=p(\sigma(T))$. So $p$ maps all points of the spectrum of $T$ to $\{0\}$, and the spectrum of $T$ is real, from which it follows that the spectrum of $T$ must be in $[0,\infty)$ because $p(s) > 0$ for $s < 0$. That the spectrum of the selfadjoint operator $T$ is contained in $[0,\infty)$ is equivalent to $\langle Tx,x\rangle \ge 0$ for all $x\in H$.

Answer 2

The spectral mapping theorem implies

$$p(\sigma(T)) = \sigma(p(T)) = \sigma(0) = \{0\}$$

therefore $\sigma(T)$ is contained in the set od zeros of $p$.

By the assumption, $s < 0$ cannot be a zero of $p$ because $p(s) > 0$. Therefore, the zeroes of $p$ are contained in $[0, +\infty\rangle$, which implies $\sigma(T) \subseteq [0, +\infty\rangle$.

Hence, $T$ is a positive operator, meaning $\langle Tx,x\rangle \ge 0, \forall x \in H$.

Answer 3

For a given $x\in H$, consider the restriction $T|_S$ to the subspace $S:=\mathrm{span}(x, Tx, T^2x, \dots T^{n-1}x) $.
Since $p(T) =0$, we have $T^nx\in S$, so $S$ is $T$-invariant.

Now, the minimal polynomial of $T|_S$ divides $p$, so it has only nonnegative real roots, hence $T|_S$ is positive semidefinite, and in particular $\langle Tx, x\rangle \ge0$.