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I'm trying to get some intuition behind the following theorem:

Let V be an F-vector space, dim $V = n$, and let $f : V \to V$ be a nilpotent endomorphism. Then $V$ has a basis $v_1,v_2,\ldots,v_n$ such that $f(v_j)$ is either zero or $v_{j+1}$.

My first question: why $f^{m-1}(V) \subset ker(f)$ ? (see second comment for answer)

Second: if we consider the quotient space $W = V/ker(f)$ and endomorphism $\tilde{f} : W \to W$. Why $\tilde{f}^{m-1} = 0$ ? (see third comment for answer)

Source: Linear Algebra II, Stoll (2007).

Mussé Redi
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    Pick some $v\in V$. What is $f(f^{m-1}(v))$? – mdp Aug 29 '13 at 10:42
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    Indeed, we have $f(f^{m-1}(v)) = f^m(v) = 0 \in ker(f) \implies f^{m-1}(V) \subset ker(f)$. – Mussé Redi Aug 29 '13 at 10:47
  • Choose some $w \in W$ and pick $v \in V$ that maps to $w$ under the canonical epimorphism $V \to W$. We then have $\tilde{f}^{m-1}(w) = f^{m-1}(v) = 0$. – Mussé Redi Sep 07 '13 at 09:52
  • nilpotent matrices have only zero as their eigenvalue so by Jardan canonical form, you are done – Ben Apr 05 '16 at 03:14

1 Answers1

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W is isomorphic but not equal to V/ker(f) where W=f(V). Consider the nilpotent matrix. The image of such a matrix intersects the kernel of the mapping. I am assuming this is the confusion. (In other words as I've defined W, f does not map W into W)

wfw
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  • I edited the question: see the second question. – Mussé Redi Aug 29 '13 at 12:11
  • $\tilde{f} : W \to W$ also has a kernel. $v_j \to ker(f)$ => $\tilde{f}(v_j + ker(f)) = 0 + ker(f)$ – wfw Aug 29 '13 at 12:33
  • Could you elaborate on $v_j \mapsto ker(f) \implies \tilde{f}(v_j+ker(f)) = 0 + ker(f)$ some more? Also I'm not sure if "$v_j \mapsto ker(f)$" is a correct notation (since $v_j$ is an element and $ker(f)$ a set)? – Mussé Redi Aug 29 '13 at 13:09
  • $\exists j$ such that $v_j \mapsto v_j+1$ where $v_j+1 \in ker(f) \implies \tilde{f}(v_j+ker(f)) = 0 + ker(f)$ s.t. $(v_j+ker(f)) \in W$ and $f(v_j+ker(f)) \subset ker(f) \implies \tilde{f}(v_j + ker(f)) = 0 + ker(f). $ My notation is keeping with W being a quotient group. – wfw Aug 30 '13 at 04:59
  • The kernel of $W$ contains one element --- namely the zero equivalence class $\bar{0} = {0 + k: k \in ker(V) }$. – Mussé Redi Sep 07 '13 at 10:06
  • (I meant the kernel of $\tilde{f}$ and $ker(f)$ for the zero equivalence class in the previous comment.) Furthermore, I still don't follow the steps you are making although your result "$0 + ker(f)$" seems correct. Lastly, why do you define $W$ to be a quotient group instead of a quotient space? – Mussé Redi Sep 07 '13 at 10:22