This is form Axler's Linear Algebra Done Right. Please allow me to borrow the screenshot from this Question : 8.13 Linearly independent generalized eigenvectors
From the above picture, we can see that in the proof of 8.13 the author uses the condition "V is finite-dimensional".
My question is that, if V is an infinite-dimensional space, does this conclusion still hold?
Thanks!
If $v_1,\cdots,v_m$ are generalized eigenvectors for distinct eigenvalues $\lambda_1,\cdots,\lambda_m$, then the subspace \begin{equation} W = \text{span}\{T^i v_j: i\in{\mathbb N}, j\in\{1,\cdots,m\}\} \end{equation} is finite dimensional and stable by $T$. The $v_j$'s are generalized eigenvectors of the restriction of $T$ to $W$, hence the theorem applies and they are linearly independent.
That said, the spectral properties of operators in infinite dimensional spaces are not limited to just eigenvalues, eigenvectors and generalized eigenvectors.