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

1. Sep 11, 2016

### Portuga

1. The problem statement, all variables and given/known data
If A and B are square matrices of same order, prove of find a counter example that if AB = 0 then BA = 0.

2. Relevant equations
$$A^{-1} A = I_n, ABC = (AB)C$$

3. 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...

2. Sep 11, 2016

### Ray Vickson

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

3. Sep 11, 2016

### 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?

4. Sep 11, 2016

### Ray Vickson

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

5. Sep 11, 2016

### Ray Vickson

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.

6. Sep 11, 2016

### PeroK

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

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

7. Sep 11, 2016

### Portuga

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

8. Sep 11, 2016

### SammyS

Staff Emeritus
Let us know what you come up with.