Counterexample to a formula for the norm of a self-adjoint operator on a Hilbert space

Question

In my functional analysis class, I have come across this exercise

For a Hilbert space $H$ and $A$ a self-adjoint operator on $H$ one has the familiar identity $$\|A\|=\sup_{\|x\|=1} \lvert \langle Ax,x \rangle \rvert $$ we are asked to show a counterexample to this when $A$ is not self-adjoint.

I cannot think of any counterexample to this. All help is appreciated.

Answer 1

On $\mathbb R^{2}$ define $A(x,y)=(-y,x)$. Then $\|A\|=1$ and $\langle Av, v \rangle=0$ for all $v$.

Answer 2

Hint: Take $H = \Bbb C^2$ with the standard inner product, and take $A$ to be the transformation associated with an upper triangular matrix.

Note that the set $\{x \in \Bbb C^2: \|x\| = 1\}$ can be parameterized as $$ x(\theta, \phi, \psi) = e^{i \psi}(e^{i \phi}\cos \theta,\sin \theta). $$ Thus, one simply needs to find the maximum $\sup_{\theta,\phi,\psi \in \Bbb R}|\langle Ax(\theta,\phi,\psi),x(\theta,\phi,\psi)\rangle|$.

---

Concretely: take $$ A = \pmatrix{0&1\\0&0}. $$ Verify that $\|A\| = 1$ by finding the eigenvalues of $A^*A$. On the other hand, for $x$ in terms of $\theta, \phi, \psi$ as above, we have $$ \begin{align} \langle x, Ax \rangle &= \langle e^{i \psi}(e^{i \phi} \cos \theta, \sin \theta), e^{i \psi}(\sin \theta,0)\rangle \\ & = \langle(e^{i \phi} \cos \theta, \sin \theta), (\sin \theta,0)\rangle \\ & = e^{i \phi} \cos \theta \sin \theta = \frac 12 e^{i \phi} \sin (2 \theta). \end{align} $$ From there, it is clear that $|\langle x, Ax \rangle| \leq 1/2 < 1$.