Find $y\in {[x]}^{\perp}$ with $\|y\| = 1$ and $\langle T^*x, y\rangle = 0$

Question

Let $\mathcal{H}$ be a complex Hilbert space with $\dim \mathcal{H}\geq 3$ and $T\in \mathbb{B}(\mathcal{H})$ such that $\langle Tx, y\rangle = 0$ for all $x\in \mathcal{H}$ and $y\in {[x]}^{\perp}$.

Let $x\in \mathcal{H}$ and $\|x\| = 1$. Why, we can choose $y\in {[x]}^{\perp}$ with $\|y\| = 1$ and $\langle T^*x, y\rangle = 0$?.

My question is the beginning of the proof of Theorem 2.7.

Answer 1

The Hilbert space has dimension $\ge 3$, hence if $u, v\in {\cal H}$, one has $\operatorname{dim}\left([u]^\perp\cap[v]^\perp\right) \ge 1$. Apply this to $x$ and $T^* x$.

An algorithm to find such an element is to take an element $w\not \in \operatorname{Span(u, v)}$, and consider $w - p(w)$ where $p$ is the orthogonal projection on $\operatorname{Span}(u, v)$. Then divide this element by its norm to get $y$.