Prove that $\{e_n\}_{n \geq 0}$ is an orthonormal family.

Question

let $H$ be a complex hilbert space, $A : H \to H$ an isometry operator such that $A(H) \neq H$, $e_0$ a unit vector that's orthogonal to $A(H)$, $e_n = A^n(e_0) \,\,\forall n \geq 1$

normality is pretty straightforward since : $\|e_n\| = \|A(A^{n-1}(e_0))\| = \|e_{n-1}\| = \dots= \|e_0\| = 1$

for orthogonality, we have by definition of $e_0$ that : $e_0 \perp e_n$

this is the part where I'm stuck, without loss of generality assume that $n > m > 0$, then :

$\langle e_n, e_m \rangle =\langle A^{n-m}(e_m), e_m \rangle$, since the adjoint operator's range is not necessarily within $A(H)$, I don't know how to proceed from here.

Answer 1

I suppose that an operator is supposed to mean a bounded linear map.

By the polarization identity any isometric linear map preserves inner products also. Hence $ \langle A^{n}e_0, A^{m}e_0\rangle=\langle A^{n-m}e_0, e_0\rangle=0$ for $n >m$.

Ref. https://en.wikipedia.org/wiki/Polarization_identity