# The sequence $\{\ker(f-\lambda Id)^{k} - \ker(f-\lambda Id)^{k-1} \}_{k}$ is decreasing

Given $$V$$ a vector space over a field $$\mathbb{K}$$ and a string $$[d_{1} < d_{2} < \cdots < d_{k} = n]$$, where $$d_{i} = \dim \ker(f-\lambda Id)^{i}$$.

I know that $$\{0\} \subset \ker(f-\lambda Id) \subset \cdots \subset \ker(f-\lambda Id)^{k} = V.$$

I would like to prove that the sequence $$\{dim \ker(f-\lambda Id)^{k} - dim \ker(f-\lambda Id)^{k-1} \}_{k}$$ Is decreasing.

Any help or solution would be appreciated.

• For subspaces $U,W \subset V$, what do you mean by $U - W$? What does it mean for a sequence of sets to be "decreasing"? – Omnomnomnom Jan 1 at 19:43
• You mean, to prove that $\{\dim\ker(f-\lambda Id)^{k} - \dim\ker(f-\lambda Id)^{k-1} \}$ is decreasing? The hint is that $\dim\ker(f-\lambda Id)^{k+1}+\dim\ker(f-\lambda Id)^{k-1} - 2\dim\ker(f-\lambda Id)^{k} = N_\lambda(k)$, where $N_ \lambda(k)$ is denoted as the number of the $k\times k$ Jordan blocks with the eigenvalue $\lambda$. – Tamshin Dion Jan 6 at 4:04
• @Omnomnomnom Yes, I forgot to write dimension, I'm going to correct that omission – jacopoburelli Jan 6 at 7:47