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I'm struggling with this exercise:

Let $n \geq 2$, $f \in End(\mathbb{C}^n)$ and $\lambda \in \mathbb{C}$. Show that if there exists an integer $k \geq 2$ such that $dim(Ker(f - \lambda id)^k) = k \cdot dim(Ker(f-\lambda id)$ then for every integer $1 \leq h \leq k$ it must be $dim(Ker(f-\lambda id)^h) = h \cdot dim(Ker(f- \lambda id)$

My thought so far: I can't see any direct proof of this fact, so I'm going to prove this by contradiction. Let $d_h = dim(Ker(f-\lambda id)^h)$. If there exists an integer $1 \leq h \leq k$ such that $d_h\neq h\cdot dim(Ker(f-\lambda id) = h \cdot d_1$ then it could be: $d_h > h \cdot d_1$ or $d_h < h \cdot d_1$, here is where I stuck: I would arrive to show that if one of the two instances happen then $d_k > kd_1$ or $d_k < kd_1$(cleary absurd because $d_k = k\cdot d_1$ for hp).

Can you give me a hint please?

Thanks. English is not my mother tongue, please excuse any errors on my part.

Ps: do you have more exercises like this? Ane references would be highly appreciated.

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    For some basic documentation on the subject (trying to avoid Jordan form, diagonalization and other hard stuff, try this in your favorite search engine : "iterated kernels", "Fitting", "decreasingly increasing kernels dimensions"... This lets you know that the case $d_h>h.d_1$ is impossible. – 2017-02-17
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    Thanks for the comment, I'm studying the "hard stuff" so I'm searching for hard exercises on Jordan Normal Form. – 2017-02-23

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You're right, it is about Jordan normal forms.

The basic observation is the following. Consider the Jordan block of size $t$ $$ B_{t} = \begin{bmatrix} \lambda & 1 & 0 & \dots & 0 & 0\\ 0 & \lambda & 1 & \dots & 0& 0\\ & & & \ddots\\ 0& 0 & 0 & \dots & \lambda & 1 \\ 0& 0 & 0 & \dots & 0 &\lambda \\ \end{bmatrix}. $$ Then $$ \dim(\ker((B_{t} - \lambda I)^{s})) = \begin{cases} s & \text{for $s \le t$}\\ t & \text{for $s \ge t$}\\ \end{cases} $$

Now the condition $$\dim(\ker(f - \lambda I)^k) = k \cdot \dim(\ker(f-\lambda I))$$ tells you that all the $\dim(\ker(f-\lambda I))$ Jordan blocks relative to the eigenvalue $\lambda$ have size $\ge k$. The result should now follow easily.