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!
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!
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:
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 $A$.
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?