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How could I show that $\forall n \ge 2$ if $A^n=0$ and $A^{n-1} \ne 0$ then $A$ has no square root? That is there is no $B$ such that $B^2=A$. Both matrices are $n \times n$.

Thank you.

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I found this post while trying to write solutions for my class's homework assignment. I had asked them precisely this question and couldn't figure it out myself. One of my students actually found a very simple solution, so I wanted to share it here:

Suppose that $N = A^2$. Since $N^n = 0$, we have $A^{2n} = 0$. Thus $A$ is nilpotent, so in fact $A^n = 0$. [This is because of the standard fact that if $P$ is any $n \times n$ nilpotent matrix, then $P^n = 0$.] Now observe that $N^{n-1} = (A^2)^{n-1} = A^{2n-2}$. But if $n \geq 2$, then $2n - 2 = n + (n - 2) \geq n$, so $N^{n-1} = 0$. This is a contradiction, so no such $A$ can exist.

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  • $\begingroup$ I would add to the explanation, "because any $(n\times n)$ nilpotent matrix $A$ satisfies $A^n = 0$." $\endgroup$ Jun 26, 2018 at 16:07
  • $\begingroup$ Yes, I am using that fact which is one of the first things one learns about niplotent matrices. I will revise it as you suggest. $\endgroup$ Jun 27, 2018 at 17:08
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Hmm. This is deeper than I thought when I gave my first glib answer.

Consider the rank $r_k$ of the matrix $A^k$. It's certainly no greater than $r_{k-1}$. So the ranks form a non-increasing sequence that looks like $$ n = r_0 \ge r_1 \ge r_2 \ge \ldots \ge r_{n-1} > r_n = 0. $$

Each matrix $A^k$ defines a linear transformation on $R^n$ whose image

$$V_k = \{ A^k {\mathbf x} : {\mathbf x} \in R^n \} $$

is a subspace of $R^n$. Furthermore, these images are nested: $$ \{0\} = V_n \subset V_{n-1} \subset V_{n-2} \subset \ldots \subset V_1 \subset V_0 = R^n$$

because if ${\mathbf u} = A^k {\mathbf x}$ is in $V_k$, then ${\mathbf u}$ can be rewritten as $$ {\mathbf u} = A^{k-1} (A^k {\mathbf x}), $$ which is clearly in the image $A^{k-1}$, i.e., $V_{k-1}$.

Suppose that for some $k$, we have $r_k = r_{k-1}$. Then by basic facts about dimensions, we have $V_k = V_{k-1}$ as well. And that means that for $j > k$, we have $V_j = V_k$, i.e., the images "stabilize". If they stablize at anything above dimension $0$, then $A$ is not nilpotent.

That means that in the sequence of non-increasing $r$-values, we can replace the weak inequality with strong inequality, to get

$n = r_0 > r_1 > r_2 > \ldots > r_{n-1} > r_n = 0$.

Since $r_1 < n$, we must have that each successive $r$ is one less than the previous one, i.e., $r_j = n-j$.

Summary so far: For an $n \times n$ matrix $A$ with $A^n = 0$ but $A^{n-1} \ne 0$, the rank of $A^k$ is $n-k$, for $k = 1, \ldots n$.

Now suppose that $A = B^2$. The sequence of matrices $B^k$ can be analyzed in the same way as the powers of $A$. Each successive matrix must have a lower rank than the last, with the rank of matrix $B^{2(n-1)} = A^{n-1}$ being at least 1. But the rank of this matrix can be no greater than $n - (2(n-1)) = 1-n$. So we have that this rank -- call it $s$ --- satisfies $1-n \ge s \ge 1$, so $1-n \ge 1$, hence $n \le 0$.

In other words, if $A$ had a square root, it would have to be a very small matrix. :)

Your problem is stated only for $n \ge 2$, but my proof seems to show that it works for $n = 1$ as well, which is wrong: for $n =1$, the matrix $[0]$ actually has the required property: its first power is zero, but its zeroth power is nonzero, and it is the square of a matrix $B$, namely $[0]$. Probably I've got an off-by-one error in my proof, or I assumed $n > 1$ somewhere without noting it.

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  • $\begingroup$ Thanks, John. You answer really helped me see more things that I should have myself. $\endgroup$
    – Wulfgang
    Nov 27, 2013 at 23:21
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So $B^{2n} = 0$, but $B^{2n-2} \ne 0$. So the Jordon Canonical Form of $B$ must contain a Jordon block whose size is at least $(2n-1)\times(2n-1)$. And since $n \ge 2$, there isn't enough room.

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I believe that what you're trying to prove is false.

Let $A = \matrix{0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0}$. Then $A^2 = 0$ but $A^1 \ne 0$. But $A$ is the square of the matrix

$$ B = \matrix{0 & 1 & 0\\0 & 0 & 1 \\ 0 & 0 & 0}. $$

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  • $\begingroup$ Wow! Thanks. This problem was on some old qualifying exam! I shall look more into this one. $\endgroup$
    – Wulfgang
    Nov 27, 2013 at 17:32
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    $\begingroup$ @ Wulfgang et al: this answer does not meet the hypotheses of the question; with $A$ $3 \times 3$, we should have $A^3 = 0$, which we do, and $A^2 \ne 0$, which we don't! $\endgroup$ Nov 27, 2013 at 17:49
  • $\begingroup$ @RobertLewis: You're completely right. I saw the "n x n" condition, and thought "Oh, it's saying the matrices are square," without noticing that the $n$ was also used in the problem setup. D'oh! $\endgroup$ Nov 27, 2013 at 18:31

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