Eigenvalues and Eigenvectors of Selfadjoint Operators

Question

I am trying to show the following:

Let $H$ be a Hilbert space. Suppose that $\|Tx\| = \|T\|$ for some unit vector $x \in H$ and for some bounded self-adjoint operator T on H. Then x is an eigenvector for $T^2$ with corresponding eigenvalue $\|T\|^2\; (= \|T^2\|)$. Moreover, either $Tx = \|T\| x$ or $Tx = \|T\| y$, where $y = \|T\| x - Tx$.

Answer 1

$T^2$ is also self adjoint and so, $$ \sup_{\|v\|=1}|(T^2v,v)|=\sup_{|v|=1}\|T^2v\| $$

The $x$ in your problem is a solution to this optimization problem. Incidentally, $\|T^2\|=\|T\|^2$.

Set $\lambda=\|T\|$. It follows that \begin{aligned} 0&\leq \|T^2x-\lambda^2x\|^2=\|T^2x\|^2-2\lambda^2(T^2x,x)+\lambda^4\|x\|^2\\ &\leq \|T^2\|^2 -2\lambda^4+\lambda^4\\ &=\lambda^4-\lambda^4=0 \end{aligned}

The conclusion to your problem should follow from looking at $$ 0=(T^2-\lambda^2 I)x=(T-\lambda I)(T+\lambda I)x $$

Answer 2

We can use the Cauchy-Schwarz inequality:

$$\|T\|^2 = \|Tx\|^2 = |\langle Tx,Tx\rangle| = |\langle T^2x,x\rangle| \stackrel{CS}{\le} \|T^2x\|\|x\| \le \|T\|^2\|x\|^2 = \|T\|^2$$

The equality condition implies there exists $\alpha \in \mathbb{C}$ such that $T^2x = \alpha x$.

We have $$\|T\|^2 = \|Tx\|^2 = \langle Tx,Tx\rangle =\langle T^2x,x\rangle = \langle \alpha x,x\rangle = \alpha\|x\|^2 = \alpha$$ so $T^2x = \|T\|^2x$.