Let $T$ be a Hermitian operator and $\|v\|=1$, prove $\langle T^2(v),v\rangle \ge\langle T(v),v\rangle^2$

Question

Let $T$ be a Hermitian operator on an inner product space $V$ over the field of complex numbers, $C$ (of a finite dimension), also assume $\|v\|=1$. Prove:

$$\langle T^2(v),v\rangle \ge\langle T(v),v\rangle^2$$

Answer 1

Edit: As pointed out by Robert Israel, this follows immediately from Cauchy-Schwarz. I so wanted to use your (wrong and now deleted hint), that I actually basically reproved Cauchy-Schwarz...

Develop the following $$ ((T-\alpha I)^2v,v)=\alpha^2\|v\|^2-2\alpha(Tv,v)+(T^2v,v). $$ Note that for every $\alpha\in \mathbb{R}$, the operator $(T-\alpha I)^2=(T-\alpha I)^*(T-\alpha I) $ is hermitian positive so $$ ((T-\alpha I)^2v,v)\geq 0. $$

We have a quadratic in $\alpha$ which is always nonnegative.

This means that its discriminant is nonpositive: $$ \Delta=4((Tv,v)^2-(T^2v,v)\|v\|^2)\leq 0. $$ Hence $$ (Tv,v)^2\leq (T^2v,v)\|v\|^2. $$

So your inequality holds for all $\|v\|\leq 1$.

Answer 2

It's not true as stated, because it has the wrong scaling behaviour with respect to $v$. Were you supposed to assume $\|v\|=1$?

EDIT: With that assumption, the hint is overkill. The result follows directly from Cauchy-Schwarz. Note that $\langle T^2 v, v \rangle = \langle T v, T v \rangle$ since $T$ is Hermitian.