2

I wanted to prove/disprove whether for every matrix $A\in \operatorname{Mat}_{n \times n}$

$\operatorname{rank} (A^2) = \operatorname{rank}(A)$ for any matrix $A\in \operatorname{Mat}_{n \times n}$

I know that for two matrices $A, B$ (it actually doesn't matters what is the size of $B$ as long as $AB$ is defined)

$$\operatorname{rank}(AB) \leq \min \{ \operatorname{rank}(A), \operatorname{rank}(B)\}$$

Therefore, $$\operatorname{rank}(A^2) \leq \operatorname{rank} (A)$$

In order to prove that:

$$\operatorname{rank}(A^2) = \operatorname{rank}(A)$$

you'd have to prove that: $$\operatorname{rank}(A^2) \geq \operatorname{rank}(A)$$ which I don't know how to prove, or finding a different way to prove directly both directions (meaning; one direction is $\leq$ and the other is $\geq$)

Arnaud Mortier
  • 27,276
Noa Even
  • 2,801

2 Answers2

8

It is not true that the square of a matrix has the same rank as the matrix itself. For instance, the $2\times 2$ matrix $A= \left[ {\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} } \right]$ has rank $1$ and satisfies $A^2=0$.

Suzet
  • 5,482
  • 12
  • 35
  • but here's what I'm confused about - for a squared matrix, $\text rank (A) = \text rank (A^T)$, so $\text rank (AA^T) = \text rank (A^2)$ – Noa Even Aug 05 '18 at 15:20
  • 2
    @Jneven This is a totally separate problem, in general $A^T$ has nothing to do with powers of $A$. I suggest you look at this then: https://math.stackexchange.com/questions/2315/is-the-rank-of-a-matrix-the-same-of-its-transpose-if-yes-how-can-i-prove-it – Suzet Aug 05 '18 at 15:21
  • but $\text rank (A) \leq \text {min}$ {${sp{R(A)}}, {sp{C(A)}}$ – Noa Even Aug 05 '18 at 15:32
  • @Jneven The rank is not multiplicative. It is not true that $\operatorname{rank}(AB)=\operatorname{rank}(A)\operatorname{rank}(B)$ as you seem to suggest. The quantities $\operatorname{rank}(AA^T)$ and $\operatorname{rank}(A^2)$ are not equal. The matrix $A$ I exhibed in my answer is a counter example for it. I let you compute $AA^T$ and see that it is not zero, whereas $A^2=0$ ! – Suzet Aug 05 '18 at 15:38
  • as my approach for proving this would be:
    1. $\text rank(A) = \text rank(A^T)$

    2 .this follows that - $\text rank(A^2) = \text rank(AA^T)$

    and according to - https://math.stackexchange.com/questions/349738/prove-rank-ata-rank-a-for-any-a-m-times-n $ \text rank(AA^T) = \text rank(A^2) = \text rank (A) $

     – Noa Even Aug 05 '18 at 15:42
  • @Jneven Your point 2. is wrong. – Suzet Aug 05 '18 at 15:43
  • notice that in the question i wrote: $\operatorname{rank}(AB) \leq \min { \operatorname{rank}(A), \operatorname{rank}(B)}$ – Noa Even Aug 05 '18 at 15:44
  • 1
    Yes, and this is true. However, $\operatorname{rank}(AA^T)=\operatorname{rank}(A^2)$ is not true. All you can say, using your inequality, is $\operatorname{rank}(AA^T)\leq \operatorname{rank}(A)$ (and also $\operatorname{rank}(A^2)\leq \operatorname{rank}(A)$). – Suzet Aug 05 '18 at 15:46
  • okay, I get it now. thanks! – Noa Even Aug 05 '18 at 16:06
6

First, it makes no sense to write $A^2$ if $A$ is not a square matrix. So you need $m=n$.

This being said, there is a class of matrices called nilpotent, from which it is trivial to build counterexamples:

$$\operatorname{rk}\left(\matrix{0&1\\0&0}\right)=1$$$$\operatorname{rk}\left(\matrix{0&1\\0&0}\right)^2=0$$

Arnaud Mortier
  • 27,276
  • I would have to disagree about the statement that "it would make no sense". in fact, I think writing $AA$ would just be weird. – Noa Even Aug 05 '18 at 15:18
  • 3
    @Jneven No, seriously, it would make no mathematical sense. A product of matrices is defined when the number of columns of the left factor is the number of rows of the right factor. – Arnaud Mortier Aug 05 '18 at 15:19
  • you're right, I understood my confusion. – Noa Even Aug 05 '18 at 15:22