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I'm working on the following exercise:

Suppose $N \in M_n(\mathbb R)$ is nilpotent and $\dim \ker N=k$, $1\leq k\leq n-1$. Show that $\dim \ker N^l\leq kl$ for every $l\geq 1$.

I have proved that this is true for $l=1, 2, 3$ because I know that for every $l\geq 1$, $\dim N^l\leq \dim N^{l+1} +\dim N^{l-1}$. I tried to proceed by induction but it didn't work.

  • please see this question, and see if any hints are available: http://math.stackexchange.com/questions/1447591/nilpotent-matrix-and-relation-between-its-powers-and-dimension-of-kernels – Sarvesh Ravichandran Iyer Oct 08 '16 at 05:30

1 Answers1

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Clearly this is true for $l = 1$.

Suppose that we know that $\dim \ker N^l \leq kl$ for some $l \geq 1$, and consider $N^{l+1}$. A vector $v$ is in $\ker N^{l+1}$ iff $N^l(v) \in \ker N$. In other words, $$\ker N^{l+1} = (N^l)^{-1}(\ker N)$$ It is a well-known fact that for any linear transformation $T: V \to W$ between finite dimensional vector spaces and any subspace $U \subseteq W$, $\dim T^{-1}(U) \leq \dim \ker T + \dim U$. Thus, we have that $$\dim \ker N^{l+1} \leq \dim \ker N^l + \dim \ker N \leq kl + k = k(l+1)$$ which was to be shown.

Alex G.
  • 9,178
  • I can prove the claim that $\dim T^{-1}(U) \leq \dim \ker T + \dim U$ if you wish, but I would encourage you to think about it on your own first. – Alex G. Oct 08 '16 at 05:40