When does an operator act diagonalisably on $V_{\lambda}$

Question

Suppose I have an operator $H$ on a complex vector space $V$.

Let $V_{\lambda}$ denote the generalised eigenspace (with respect to $H$) of a maximal eigenvalue $\lambda$ (maximal of real value). (Maximality may not come into the question here, but it's part of the problem I'm trying to solve)

Then it is clear that $HV_{\lambda} \subset V_{\lambda}$, and we can ask when does $H$ acts diagonalisably on $V_{\lambda}$.

If $H$ does act diagonalisably on $V_{\lambda}$, it must be with eigenvalues $\lambda$ because if $\{v_1,..,v_m\}$ denotes a basis of $V_{\lambda}$ then for $1 \leq i \leq m$, $(H - \lambda)^{k_i}v_i = 0$ for some natural number $k_i$. If we assume $Hv_i = a_i v_i$ then $(a_i - \lambda)^{k_i} = 0$ and so $a_i = \lambda$.

Thus to say that $H$ acts diagonalisably on $V_{\lambda}$ is to say that every vector in $V_{\lambda}$ is an eigenvector with respect to $\lambda$.

Questions:

1. is the above analysis correct?

2. is there in general a canonical basis of $V_{\lambda}$?

About the second question; if the basis of $V_{\lambda}$ is the eigenvectors of $H$ with respect to $\lambda$ then it becomes trivial that $H$ acts diagonalisably on $V_{\lambda}$, which isn't the case most of the time, so this shouldn't be correct..

Thanks for any assistance!

Answer 1

Your analysis is correct. The fact that $\lambda$ is the maximal eigenvalue is irrelevant.

Finding a basis of $V_\lambda$ is a subproblem of the problem of computing the Jordan form of a matrix.