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.
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.