If $x,Tx,T^2 x,...,T^{n-1} x$ spans a finite-dimensional Hilbert $H$, then there exists an orthonormal basis consisting of eigenvectors of $T$

Question

Let $H$ be a finite-dim inner product space spanned by $x,Tx,...,T^{n-1} x$ where $T$ is a linear operator on $H$, then does there always exist an orthonormal basis consisting of eigenvectors of $T$? We may assume that $x,Tx,...,T^{n-1} x$ are linearly independent if necessary.

Answer 1

If for some $x\in H$ set $\{x,Tx,\ldots,T^{n-1}x\}$ forms basis in $H$, then $x$ is a cyclic vector of a linear transformation $T$. A linear transformation has a cyclic vector if and only if its minimal polynomial is the same as characteristic polynomial.

On the other hand, we know that a linear transformation has a basis of eigenvectors if and only if it's diagonalizable (precisely in this basis), that happens when minimal polynomial splits into distinct linear factors.

However, it isn't hard to come up with a linear transformation whose matrix has a minimal polynomial equal to characteristic, but it doesn't split into distinct linear factors. Think about nilpotent matrices for example.

Answer 2

Here's how I go about tackling questions like this. First, I notice that the claim seems highly plausible in the case that $T$ is, say, full-rank and self-adjoint. So I search my zoo of "usual" bad matrices and try a few cases that are less plausible---for instance, take $H=\mathbb{R}^2$ with the usual dot product, $T = \begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$, and $x = \begin{bmatrix}0\\1\end{bmatrix}$...