If $A^m = 0$, then $\mbox{rank}(A) \leq \frac{m-1}{m}{n}$

Question

Let $A$ be a $n \times n$ real matrix. Show that if $A^m = 0$, then $\mbox{rank}(A) \leq \frac{m-1}{m}{n}$

My attempt:
If $m=1$, then $A=0$ so $\mbox{rank}(A)=0$.
If $m=2$, we have $\mbox{im}(A) \subset \ker(A)$ so $2\operatorname{rank}(A) \leq \dim \mbox{im}(A) + \dim \ker(A)=n$
For arbitrary $m$, I want to use induction. $B=A|_{\mbox{im}(A)}$ satisfies $B^{m-1}=0$ so $\mbox{rank}(B) \leq \frac{m-2}{m-1}\mbox{rank}(A)$. Thus $\dim \ker B \geq \mbox{rank}(A)- \frac{m-2}{m-1}\mbox{rank}(A) =\frac{1}{m-1}\mbox{rank}(A)$ by rank-nullity theorem. Thus $n=\dim \ker A + \mbox{rank}(A) \geq \dim \ker B +\mbox{rank}(A)\geq \frac m {m-1}\mbox{rank}(A)$

Is this ok?

Answer 1

I would rather prove this quite differently.

Below is the Sylvester's Rank Inequality: $$r(A)+r(B) \le n + r(AB)$$ It could be generalized by induction as: $$\sum_{k=1}^m r(A_k) \le n(m-1) + r\left(\prod_{k=1}^m A_k\right)$$

Taking $A_1=A_2=...=A_m = A$, we will get $$m\cdot r(A) \le n(m-1) + r(A^m) \implies \boxed{r(A) \le \frac{m-1}{m}n}$$ as desired

Answer 2

This is essentially correct. You should clarify that by $B=\left.A\right\rvert_{\operatorname{im}A}$ you mean the linear map $\operatorname{im}A\to\operatorname{im} A$ and not the map $\operatorname{im}A\to\Bbb R^n$. Depending on how much the person whom you are speaking to values your ability to fill in details, you might want to address the fact that:

1. your inductive hypothesis is that the inequality holds $(\forall k<m,\forall n,\forall A,\cdots)$ as opposed to, say, $n$ being fixed.

2. the inductive hypothesis speaks of matrices, but you have chosen the notation in the inductive step to speak of linear maps to a substantial length. This may be addressed easily in three ways that I can think of: (a) saying that in this specific instance the problem may be reformulated matricially; (b) describing the matrix of $B$ in some basis; (c) starting all over again by saying that you are considering the linear-map formulation of the problem.

Answer 3

Here's a solution using a Jordan canonical form for $A$ . . .

Let $A$ be a nonzero $n{\,\times\,}n$ matrix such that $A^m=0$ for some positive integer $m$, and let $r=\text{rank}(A)$.

Fix a Jordan canonical form for $A$.

Since $A$ is nonzero, there is at least one nonzero Jordan block.

Let $B_1,...,B_k$ be the sequence of nonzero Jordan blocks.

Let $n_i$ be the size of $B_i$ (i.e., $B_i$ is an $n_i{\,\times\,}n_i$ matrix), and let $N=\sum_{i=1}^k n_i$.

Then $B_i$ has rank $n_i-1$, hence $r=\sum_{i=1}^k (n_i-1)=N-k$.

Since $A^m=0$, we must have $B_i^m=0$ for all $i$, hence $m\ge n_i$ for all $i$.

For each $i$, let $x_i=m-n_i$, and let $X=\sum_{i=1}^k x_i$.

Then $X=\sum_{i=1}^k (m-n_i)=km-N$, hence \begin{align*} \Bigl(\frac{m-1}{m}\Bigr)n-r &\ge \Bigl(\frac{m-1}{m}\Bigr)N-r \\[4pt] &= \Bigl(\frac{m-1}{m}\Bigr)N-(N-k) \\[4pt] &= \frac{km-N}{m} \\[4pt] &= \frac{X}{m} \\[4pt] &\ge \;0 \\[4pt] \end{align*} so we have $ r\le\Bigl({\Large{\frac{m-1}{m}}}\Bigr)n $ as was to be shown.