2

I know it is possible to have such a matrix if $A$ is $3\times 3$. I think it won't be possible in this case but I'm not sure how to prove a general case.

A specific case proof would work something like this.

$A$ is determined by its action on $e_1,e_2$
Say $Ae_1=\alpha e_1 + \beta e_2$.
Note that $\alpha=0$ else $A^3\ne0$
Similarly $Ae_2=\gamma e_1$
Hence both $\beta=0=\gamma$

How to prove for larger dimension? That is, $A\in M_{n\times n}$ then it is not possible that $A^n\ne0=A^{n+1}$

Anvit
  • 3,379

3 Answers3

2

Consider the images of $A, A^2$ and $A^3$ as subspaces of the plane. If the image of $A$ is the whole plane, then so are the images of $A^2$ and $A^3$. If the image of $A$ is just the origin, then so is the image of $A^2$.

What remains is to study what happens when the image of $A$ is a line $\ell$. Now consider the image of $A^2$. This is equal to the image of $\ell$ under $A$. This is either all the line, or just the origin. If $A\ell$ is just the origin, then that means $A^2=0$ (one application of $A$ sends any point to $\ell$, the second application sends all those points to the origin).

If the image is all of $\ell$, then the image of $A^2$ is the same as the image of $A$, and the image of $A^3$ must again necessarily be the same $\ell$, making $A^3\neq0$.

Arthur
  • 199,419
  • Is it possible to be get a more general argument? This argument is not very scalable – Anvit Apr 26 '19 at 07:09
  • 1
    @Anvit It's very scalable. We just need a slight rephrasing. Note that we have, for a general $n\times n$ matrix $A$, a chain $$\Bbb R^n\supseteq A\Bbb R^n\supseteq A^2\Bbb R^n\supseteq \cdots \supseteq A^{n+1}\Bbb R^n$$where for each step the dimension either decreases or it doesn't. If it doesn't decrease at some step, then it doesn't decrease at any later steps either. So if $A^{n+1}R^n = {0}$, then we must have ${0}$ at one of the earlier steps as well. – Arthur Apr 26 '19 at 07:16
2

You can handle the general case using Cayley-Hamilton theorem.

  • Assume there is $A$ with $A^{n+1} =O_{n\times n}$ and $A^n \neq O_{n\times n}$

Using Cayley-Hamilton, this will lead to a contradiction as follows:

  • According to Cayley-Hamilton we have for the characteristic polynomial $p_A(t) = \sum_{k=0}^na_k t^k$ of $A$: $$p_A(A) =\sum_{k=0}^na_kA^k = O_{n\times n}$$
  • Multiplying this by $A^n$ and using $A^{n+1} = O_{n\times n}$ gives $$A^np_A(A) = a_0A^n = O_{n\times n} \stackrel{A^n \neq 0}{\Rightarrow} a_0 = 0$$
  • The same way, you get successively multiplying by $A^{n-k}, (k=0,\ldots , n-1)$ $$a_1 = \ldots = a_{n-1} = 0$$

It follows $$p_A(A) = a_n A^n = O_{n\times n} \Rightarrow A^n = O_{n\times n} \mbox{ Contradiction!}$$

trancelocation
  • 32,243
0

Start with Cayley-Hamilton for a $2 \times 2$ matrix:

$A^2 - (\text{Tr}A)A + \det A = 0; \tag 1$

multiply it by $A$:

$A^3 - (\text{Tr}A)A^2 + (\det A)A = 0; \tag 2$

use $A^3 = 0$:

$-(\text{Tr}A)A^2 + (\det A)A = 0; \tag 3$

times $A$ again:

$-(\text{Tr}A)A^3 + (\det A)A^2 = 0; \tag 4$

use $A^3 = 0$ again:

$(\det A)A^2 = 0; \tag 5$

since $A^2 \ne 0$, this yields

$\det A = 0; \tag 6$

go back to (3) using this:

$(\text{Tr}A)A^2 = 0; \tag 7$

again, since $A^2 \ne 0$:

$\text{Tr}A = 0; \tag 8$

now use (6) and (8) in (1):

$A^2 = 0 \Rightarrow \Leftarrow A^2 \ne 0; \tag 9$

this contradiction implies there is no such $A$.

Nota Bene: If the Cayley-Hamilton theorem is not available, (1) can be verified by simply grinding out the algebra in terms of the entries of $A$. Also, I was about to $\LaTeX$ up the general case but trancelocation has it covered. End of Note.

Robert Lewis
  • 71,180