If A, B are n order square matrices, and AB=0, then BA=0?

[Portuga]

Homework Statement

If A and B are square matrices of same order, prove of find a counter example that if AB = 0 then BA = 0.

Homework Equations

$$A^{-1} A = I_n, ABC = (AB)C$$

The Attempt at a Solution

$$AB = 0 \Rightarrow A^{-1} A B = A^{-1} 0 \Rightarrow (A^{-1} A) B = A^{-1} 0 \Rightarrow I_n B = A^{-1} 0 \Rightarrow B = A^{-1} 0 \Rightarrow B = 0 \Rightarrow BA = 0A = 0.$$
I am not pretty sure if this procedure really solve the problem, so I would like some advices...
Thanks in advance.

 

[Ray Vickson]

Homework Statement

If A and B are square matrices of same order, prove of find a counter example that if AB = 0 then BA = 0.

Homework Equations

$$A^{-1} A = I_n, ABC = (AB)C$$

The Attempt at a Solution

$$AB = 0 \Rightarrow A^{-1} A B = A^{-1} 0 \Rightarrow (A^{-1} A) B = A^{-1} 0 \Rightarrow I_n B = A^{-1} 0 \Rightarrow B = A^{-1} 0 \Rightarrow B = 0 \Rightarrow BA = 0A = 0.$$
I am not pretty sure if this procedure really solve the problem, so I would like some advices...
Thanks in advance.

You cannot assume that $A^{-1}$ exists. Some nonzero $n \times n$ matrices do not have inverses.

 

[Portuga]

Ok. I was using sagemath to make some reasonings. So I put there a generic 2x2 A matrix,
$$\begin{bmatrix} x & y\\ z & t \end{bmatrix}$$
and solved AB = 0 for B. The software answered that the only solution is a null matrix. That's why I am trying to prove it. Did I miss something?

 

[Ray Vickson]

Ok. I was using sagemath to make some reasonings. So I put there a generic 2x2 A matrix,
$$\begin{bmatrix} x & y\\ z & t \end{bmatrix}$$
and solved AB = 0 for B. The software answered that the only solution is a null matrix. That's why I am trying to prove it. Did I miss something?

What about
$$A = \pmatrix{1&1\\1&1}, \: B = \pmatrix{1 & -2 \\-1 & 2} ? $$
Do we have $AB = 0$? Is $A \neq 0$? Is $B \neq 0$?

 

[Ray Vickson]

What about
$$A = \pmatrix{1&1\\1&1}, \: B = \pmatrix{1 & -2 \\-1 & 2} ? $$
Do we have $AB = 0$? Is $A \neq 0$? Is $B \neq 0$?

You should AVOID using software like sagemath or whatever when you are learning the subject. In this case the answers it gives you may be highly misleading, and may cause you to "learn" things that are actually false.

 

[PeroK]

Ok. I was using sagemath to make some reasonings. So I put there a generic 2x2 A matrix,
$$\begin{bmatrix} x & y\\ z & t \end{bmatrix}$$
and solved AB = 0 for B. The software answered that the only solution is a null matrix. That's why I am trying to prove it. Did I miss something?

I think what you found there was that the 0 matrix is the only matrix, $X$, for which:

For all matrices $A$ we have $AX = 0$

 

[Portuga]

Ok, I got the point. So, I should make a counter example. Thanks!

 

[SammyS]

Ok, I got the point. So, I should make a counter example. Thanks!

Let us know what you come up with.