Prove that $ \text{rank}(A^{3}) < \text{rank}(A^{2}) < \text{rank}(A) $ if $ A $ is a $ (10 \times 10) $-matrix with nilpotence equal to $ 3 $.

Question

I know that generally, for any square matrix $ B $, $ \text{rank}(B^{2}) $ is less than or equal to $ \text{rank}(B) $, but I’m having trouble with this proof.

Answer 1

You are correct to note that the inequality $\operatorname{rank}(A^2) \le \operatorname{rank}(A)$ always holds, and the inequality $\operatorname{rank}(A^3) \le \operatorname{rank}(A^2)$ also holds.

To verify this, you should check that in general, the null-space of $A^n$ contains the null-space of $A^{n+1}$ as a subspace, and apply the rank-nullity theorem.

In particular, for this problem you need to show that the inclusions of null-spaces are strict given the hypotheses. Thus, you need to show that $A^3 x = 0, A^2 x \ne 0$ for some $x$, and $A^2y = 0, Ay \ne 0$ for some $y$. The existence of such an $x$ and $y$ is left to the reader.

Answer 2

If a matrix $A$ is nilpotent (of any size and of any index of nilpotence) and $\def\rank{\operatorname{rank}}\rank A^i=\rank A^{i+1}$, then $A^i=0$.

Indeed, if $V=\mathbb R^n$, recall that the rank of $A$ is the dimension of $AV$. We have $A^iV\supseteq A^{i+1}V$ and, since both subspaces have the same dimension by hypothesis, we have $A^iV=A^{i+1}V$. But then $A^{i+2}V=AA^{i+1}V=AA^{i}V=A^{i+1}V$, and $A^{i+3}V=AA^{i+2}V=AA^{i+1}V=A^{i+2}$, and so on. We thus see that $A^jV=A^iV$ for all $j\geq i$. But $A$ is nilpotent, so there is an $k\geq i$ such that $A^kV=0$, and it follows then that in fact $A^iV=0$.

Answer 3

First note that $A^3=0$ and $A^2\neq0$ hold be the hypothesis that the order of nilpotence of $A$ is$~3$. So you only need to worry about proving $\def\rk{\operatorname{rank}}\rk(A^2)<\rk(A)$.

In general for commuting square matrices $A,B$, if it is given that $AB=BA$ has the same rank as $B$, then this means that the restriction of (the linear operator defined by) $A$ to the image (or range) subpace $W$ of $B$ is a surjective (in fact invertible) operator $W\to W$. This is so because the image $A(W)$ is the image of the composition $AB$, while the image of $BA$ is contained in the image of $B$; thus $A(W)\subseteq W$, but by the rank hypothesis $\dim(A(W))=\dim(W)$, so necessarily $A(W)=W$. One can repeat operation of the restriction of $A$ to $W$; its powers are also surjective $W\to W$, so the image of any composition $A^iB$ will still be equal to $W$.

Applying this for $B=A$, one sees that $\rk(A^2)=\rk(A)$ would imply that the images of all $A^i$ are the same for all $i\geq1$. (More generally, applying it for $B=A^k$ shows that the condition $\rk(A^{k+1})=\rk(A^k)$ implies that the images of all $A^i$ are the same for all $i\geq k$.) But this would contradict the hypotheses $A^3=0$ and $A^2\neq0$. So one must have $\rk(A^2)<\rk(A)$, as desired.