# How do you find the rank of a matrix given a different version of that same matrix?

I'm pretty sure you have to use the dimension of the column space but I can't figure this out:

If $A$ is a $3\times 3$ matrix such that $A^2 = 0$, then show that the rank of $A = 1$.

If anyone can help, thanks!

• What do you mean by "find the rank of $A=1$"? The question doesn't seem to make any sense.
– pki
Dec 5 '11 at 1:14
• This is question number 30,000 on the site, by the way. Where are the balloons? Dec 5 '11 at 1:15
• Problem was worded weird... sorry Dec 5 '11 at 1:15
• Please, don't shout. Dec 5 '11 at 1:16
• @jmendegan It makes more sense now. But maybe you want to assume that $A \neq 0$? The zero matrix of this size certainly satisfies $A^2 = 0$. Dec 5 '11 at 1:17

More generally, if an $n \times n$ matrix has rank $r$, that means that the column space of $A$ has dimension $r$, and the null space has dimension $n-r$. In order for $A^2 = 0$ it is necessary and sufficient that the column space is contained in the null space, and this can only happen if $r \le n-r$, i.e. $r \le n/2$.

As stated, the conclusion does not follow: if $$A=0$$, then $$A^2=0$$, but $$A$$ does not have rank $$1$$, it has rank $$0$$.

If we add the assumption that $$A$$ is not the zero matrix, on the other hand, then the result does follow:

1. Using the Jordan Canonical Form. Since $$A^2=0$$, the minimal polynomial of $$A$$ is either $$t$$ or $$t^2$$; it cannot be $$t$$ since we are assuming that $$A\neq 0$$, so the minimal polynomial is $$t^2$$. That means that the Jordan Canonical form of $$A$$ has a $$2\times 2$$ Jordan block. From this, it is easy to determine the rank and/or nullity of $$A$$.

2. Without Using the Jordan Canonical Form or the Minimal Polynomial. The nullspace of $$A$$ is of dimension $$1$$ or $$2$$ (it cannot be of dimension $$3$$, because $$A\neq 0$$; and it cannot be of dimension $$0$$, because then $$A$$ would be invertible, and hence $$A^2$$ would be invertible). We need to show that the dimension is $$2$$.

Since $$A^2 = 0$$, that means that the image of $$A$$ is completely contained in the nullspace of $$A$$. Hence, $$\mathrm{rank}(A)\leq \mathrm{nullity}(A)$$. By the Rank-Nullity Theorem, we know that $$3 = \mathrm{rank}(A) + \mathrm{nullity}(A)$$. So... what must $$\mathrm{nullity}(A)$$ be for everything to work out?

• Sorry... just trying to get all this straight. I'll accept an answer soon. Dec 5 '11 at 1:42