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I am interested in feedback on my attempt to write a version of this proof which I wrote up after asking this question and looking at those answers along with the original generalization of the problem. Here is my version of the proof.

$\mathbf{Problem}:$ Show that $A^{10} = 0$ given $A$ is an $10 \times 10$ matrix over $\mathbb{C}$ and $A^{10,000} = 0$

I will generalize by proving that the following theorem holds and thus our problem is solved.

$\mathbf{Theorem}:$ Given an $n \times n$ linear transformation, $A$, over a field, $K$, if $A^m = 0$ then $A^n = 0$ for all $m \geq n$

$\mathbf{Proof}$: If $m = n$ we are done. Let $m > n$. Note that $A$ is a $\mathbb{C}$-module endomorphism on a vector space over $\mathbb{C}$,$V$, with a linear transformation $A:V \rightarrow V$ and we can describe $A^m$ with a sequence of linear transformations of length $n$.

If $A$ is injective then every power of $A$ is also injective and will always have a trivial kernel hence never equal to $0$. Then $A$ must have a non-trivial kernel. By the 1st isomorphism theorem $Im(A)$ is isomorphic to a proper subgroup of $V$. And proper subgroups of $V$ have integer rank which is at most $n-1$.

Assuming $Im(A)$ has a rank of $n-1$. If $A_1:Im(A) \rightarrow Im(A)$ (an endomorphism on $Im(A)$ which is pulled directly from it's embedding in $A$) is injective than no power of $A$ will be $0$ so we again assume that $A_1$ has a non-trivial kernel. We find that $Im(A_1)$ is a proper subgroup of $Im(A)$ which must have a rank of at most $n-2$. We can continue in this way showing that while at any point our "chain" could end in zero the "worst case" scenario is that we need a chain of $n$ homomorphisms ($A^n$) to get to $0$ so in any case $A^n = 0$. Q.E.D.

$$\require{AMScd}\begin{CD} V @>{A}>> Im(A) @>{A_1}>> Im(A_1) @>{A_2}>> \cdots Im(A_{n-3}) @>{A_{n-2}}>> Im(A_{n-2}) @>{A_{n-1}}>> 0\\ \end{CD}$$

(My illustration of my argument where each $A_i$ is an embedded homomorphism inside of $A_{i-1}$ which illustrates the last map in the composition that takes $A^i(V) = W \leq V$)

I am not looking for a new solution just feedback on how to better present my understanding of how to solve this problem. I am still new to rigorous proof writing and am working on taking my "hand-wavy" explanations that make sense in broad strokes and giving them the detail and precision necessary to train to become a mathematician.

Edit: I will be fixing any errors in the proof as I find them. Pardon the dust

Algebra is Awesome
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  • One consequence of your theorem is that if $A^m=0$ then $A=0$. That’s patently false. – amd Jan 14 '20 at 07:05
  • Thank you! I was over zealous and addressing every case and forgot that the theorem need not even apply to m < n. i have made the correction – Algebra is Awesome Jan 14 '20 at 07:18
  • @AlgebraisAwesome Well, if $m<n$, then clearly also $A^m=0$ implies $A^n=0$. – Hagen von Eitzen Jan 14 '20 at 07:27
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    Your “theorem” is badly phrased, as it is unclear the scope of the “for all”. A better way would be to say: “Let $A$ be an $n\times n$ matrix. If there exists $m\geq n$ such that $A^m=0$, then $A^n=0$.” (Or better yet, as Hagen notes, simply “If there exists $m\geq 0$ such that $A^m=0$, then $A^n=0$”). As to the proof, do you know about the minimal polynomial? – Arturo Magidin Jan 14 '20 at 07:33
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    (P.S. We don’t generally say a linear transformation “is $n\times n$“; a matrix is, a linear transformation would be an operator on an $n$-dimensional space... sure, there is a way to identify a matrix with a linear transformation, but still: they correspond to each other, they aren’t the same thing. – Arturo Magidin Jan 14 '20 at 07:34

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This isn't true for an $n \times n$ matrix as originally asked. I'm leaving this answer up even though the question has been amended appropriately, because it demonstrates why.

For a vector space of dimension $n$ take a basis $a_1, \dots a_n$ and consider the linear transformation $T$ which takes $a_1\to 0$ and $a_r \to a_{r-1}$ for $r\gt 1$. Then $T^n=0$ while $T^r\neq 0$ for $r\lt n$.

Let $M$ be the matrix associated with $T$.

So you would need $n\le 10$.

To have a nilpotent matrix (finite dimension for the moment) you need it to reduce the dimension of the image space by at least $1$ with each operation until you get to zero. This example shows that you can achieve $1$ every time.

Mark Bennet
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  • Thank you! I hadn't realized I used "$ n \times n$" in the post and title but meant "10x10" as I did in my original post. Silly, I will fix it – Algebra is Awesome Jan 14 '20 at 07:20