Endomorphisms and nilpotency

Question

I tried to proof something about nilpotent matrices but I'm not quite sure if my proof is correct, it is at least a bit handwavy so I'd love if someone would explain to me how I should make it more rigorous. The statement is as follows: Let $N$ be a $12 \times 12$ matrix so that $N^4 =0$, proof that $N$ has rank at most 9.

I tried to proof it by contradiction and my argument is as follows: Consider the situation that $rk (N) >9$, considering $N$ is nillpotent, there exists a basis $B = \{b_{1}, b_{2},\dots,b_{12}\}$ such that for all $i \in \{1,2,\dots,12\}$ $N\cdot b_{i} = b_{i+1}$ or $N \cdot b_{i} = 0_{v}$. Consider this basis $B$ as a multiset, if $rk(N) >9$ then $rk(N)$ is atleast $10$, so for at most $2$ different $i \in \{1,2,\dots,12\}$ we have that $N\cdot b_{i} = b_{i+1}$ or $N \cdot b_{i} = 0_{v}$ Which means that after applying $N$ 4 times at $B$ we find that $B$ consist of at most $4$ distinct elements, hence we find that $rk(N^4) \geq 4$ but we are given that $N^4 = 0$, which lead us to a contradiction.

My biggest problem with this line of reasoning is the line where I state that for at most $2$ different $i \in \{1,2,\dots,12\}$ we have that $N\cdot b_{i} = b_{i+1}$ or $N \cdot b_{i} = 0_{v}$, I know this is true but the fact is that I don't really state anything about the other generators in $B$, this "proof" however is precisely how I tend to think about nillpotent endomorphisms, namely as linear transformations that "make your basis smaller after every application".

Any tips on how to make this argument work better would be greatly appreciated!

Answer 1

Consider the sequence of iterated kernels: $$E_0=0\subset E_1=\ker N\subset E_2=\ker N^2\subset E_3=\ker N^3 \subset E_4=\ker N^4=K^{12}$$ and let $d_i=\dim E_i,\quad i=0,1,2,3,4$. We have $$d_0=0\le d_1\le d_2\le d_3\le d_4=12.$$ Furthermore, for $1\le i\le 3$, we have an injective homomorphism \begin{align*} E_{i+1}/E_i&\longrightarrow E_i/E_{i-1}\\ v&\longmapsto N\cdot v \end{align*} which proves $\;d_{i+1}-d_i\le d_i-d_{i-1}$.

Now suppose $\; \dim E_1\le 2$. Then by the previous observation,$\;\dim E_2\le 4,\enspace \dim E_3\le 6$ and $\;\dim E_4\le 8$, wheras we know it's $12$.

So $\;\dim \ker N\ge 3$ and, by the rank-nullity theorem, $\operatorname{rank}N\ge 12-3=9$.

Answer 2

The rank of $N$ is the dimension of the image of the endomorphism $f_N$ associated with $N$. The rank plus the dimension of $\ker f_N$ equals 12, so you must show that $\dim \ker f_N \ge 3$.

Take a non-zero vector $v_1$. Put $v_2 = f_N(v_1)$, $v_3 = f_N(v_2)$ and $v_4=f_N(v_3)$. By hypothesis $f_N(v_4)=0$. At some point in the sequence $v_1, ... , v_4$, $v_i \ne 0$ and $f_N(v_i) = 0$.

Then select $w_1$ not in the linear subspace spanned by $v_1, ... , v_4$ and repeat. And finally select $t_1$ not in the linear subspace spanned by the $v$ and the $w$.

You will find three linearly independent vectors in $\ker f_N$. $\square$