# A manipulative matrix question

1. May 12, 2015

### Raghav Gupta

1. The problem statement, all variables and given/known data

If A and B are 2 matrices such that AB = A and BA =B, then B2 is equal to
B
A
Zero matrix
I
2. Relevant equations
We can pre or post multiply a matrix on both sides of equation.

3. The attempt at a solution
(AB).(BA) = A.B
AB2A = A.B
Pre multiply both sides by A-1
We get B2A = AB
Implies B2AB = AB as (A = AB)
Post multiplying by B-1 and A we get B2 = I

Is this correct?

2. May 12, 2015

### SammyS

Staff Emeritus
Shouldn't that last equation be as follows?

B2A = B

By the way: Are you certain that matrix, A, has an inverse?
Are you certain that matrix, A, has an inverse?

3. May 12, 2015

### Staff: Mentor

I don't see anything in the problem that says that either matrix is invertible, so there's no justification for assuming that A-1 exists.

4. May 12, 2015

### scientific601

Maybe it was a trick question - nothing involving inverse matrices. Both A and B could be identity matrices. This would satisfy the problem statement in the strictest sense.

5. May 12, 2015

### SammyS

Staff Emeritus
I started with A2 .

6. May 12, 2015

### Staff: Mentor

That seems the long way around if the goal it to find B2, but then I don't know what you did.

7. May 12, 2015

### SammyS

Staff Emeritus
Well, I suppose we'll have to see what OP does before we can all compare notes.

8. May 12, 2015

If you start with $B^2$,
Then i got A=I.
$B^2=B.B=B.BA=B^2A$
$B^2=B^2A$
A=I satisfies the above equation.
And if i put it on AB=A, then i got B=I.

9. May 12, 2015

### SammyS

Staff Emeritus
While I satisfied that equation, it's not necessary that A = I .

10. May 13, 2015

### Staff: Mentor

As Sammy said, B isn't necessarily I. Here's a counterexample, using your assumption that A = I.
$$B = \begin{bmatrix} 1 & 0 \\ 0 & 0\end{bmatrix}$$
This matrix satisfies B2 = B2A = B2I, yet $B \neq I$.

11. May 13, 2015

### Raghav Gupta

So,
I started with B2 ,

$B^2 = B.B = (BA).(BA) = B(AB)A = (BA)A= BA=B$ Is multiplication of matrices associative which I am doing here?

Starting with A2.
Following the above similar pattern , I get A2= A.
How did you Sammy get B2 value by starting from A2?

12. May 13, 2015

### Dick

Yes, multiplication of matrices is certainly associative.

13. May 13, 2015

### Raghav Gupta

How?

14. May 13, 2015

### Dick

15. May 13, 2015

### Raghav Gupta

16. May 13, 2015

### Dick

It's not deep, you just have to pay attention to the indices. Ignore the ring stuff. Just use that for real numbers (ab)c=a(bc).

17. May 13, 2015

### Raghav Gupta

Thanks, got it.

18. May 13, 2015

### SammyS

Staff Emeritus
Of course, I didn't follow that pattern, but, yes, A2= A.

A2 = A2

BAAB = BAAB

((BA)A)B = (BA)(AB)

(BA)B = BA

BB = B

Here are two such matrices.
$\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]$

$\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]$

Last edited: May 13, 2015
19. May 13, 2015

### Raghav Gupta

How you have written A2 = BAAB ?

20. May 13, 2015

### SammyS

Staff Emeritus
Here are two such matrices.
$\displaystyle \left[\begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix}\right]$

$\displaystyle \left[\begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix}\right]$
I didn't say that.

I said

BAAB = BAAB

Now that I look at that, I only needed:

BAB = BAB

(BA)B = B(AB)

BB = BA

B2 = B