Inner product of composed function and vector

Question

I am self-studying Sheldon Axler's Linear Algebra Done Right. One of the lemmas required to prove the Real Spectral Theorem contains a move I can't understand.

Suppose T $\in L(V)$ is self-adjoint, and let $\alpha, \beta \in \mathbb{R}$ are such that $\alpha^2 < 4\beta$.

The proof follows: \begin{align} \langle(T^2 + \alpha T + \beta I)v, v\rangle = \langle T^2v, v\rangle + \alpha\langle Tv, v \rangle + \beta \langle v, v \rangle \\ = \langle Tv, Tv \rangle + \alpha\langle Tv,v \rangle + \beta||v||^2 \end{align}

Why is $\langle T^2v, v\rangle = \langle Tv, Tv \rangle$ ?

Answer 1

Since $T$ is self-adjoint, $T^{2} = T^{*} T$ and, thus, $$\langle T^{2} v, v \rangle = \langle T^{*} T v, v \rangle = \langle Tv, TV \rangle.$$