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

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!

• What is a canonical basis? – José Carlos Santos Nov 10 '18 at 11:49
• @JoséCarlosSantos I just mean is there a good guess of what this basis could be – Mariah Nov 10 '18 at 11:51

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.