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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!

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

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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?

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

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