# Square of transpose of two matrices

1. Nov 14, 2012

### V0ODO0CH1LD

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

Let A and B be two square matrices of order n such that AB = A and BA = B. Then, what is the value of [(A + B)t]2?

2. Relevant equations

3. The attempt at a solution

[(A + B)t]2 = AtAt + AtBt + BtAt + BtBt.

I tried to use the fact that AB = A and BA = B to keep going but I didn't succeed at it..

2. Nov 14, 2012

### Yuu Suzumi

Do you remember what $(AB)^T$ is in terms of $A^T$ and $B^T$? Have you tried using that?

3. Nov 14, 2012

### V0ODO0CH1LD

yeah, the expression would look like $A^t A^t + (BA)^t + (AB)^t + B^t B^t = A^t A^t + B^t + A^t + B^t B^t$. But I have a tendency to go down paths that take me nowhere when messing around with equations and expressions.. That is where I got stuck in the first place..

4. Nov 14, 2012

### haruspex

I don't understand how you would know when you have the desired answer. But you can simplify the above further. Investigate ABA.

5. Nov 15, 2012

### V0ODO0CH1LD

(i) $(A+B)^2$; (ii) $2(A^t B^t)$; (iii) $2(A^t+B^t)$; (iv) $A^t+B^t$; (v) $A^tB^t$;

6. Nov 15, 2012

### Yuu Suzumi

With the hint given by haruspex you can arrive at one of those!

Find two ways to write $ABA$!

7. Nov 15, 2012

### V0ODO0CH1LD

$(AA)^t + B^t + A^t + (BB)^t = (ABA)^t + B^t + A^t + (BAB)^t = (AB)^t + B^t + A^t + (BA)^t = (A)^t + B^t + A^t + (B)^t = 2(A^t + B^t)$

Is that correct!?

8. Nov 15, 2012

### Yuu Suzumi

Yes. At least that's what I got too :-)

Do you know where the thread is in which they describe how much we are allowed to help im HW forums? I can't find it and therefore I am limiting myself to remarks like above.

9. Nov 15, 2012

### V0ODO0CH1LD

I didn't even know there were rules on that, oh well.. Thanks!!

10. Nov 15, 2012

### Dick

https://www.physicsforums.com/showthread.php?t=414380 Solving the problem for them is right out. I'd say give them the smallest clue you think will get them heading in the right direction. The hints you gave so far worked fine, I wouldn't do more even if it is tempting to do so.

11. Nov 16, 2012

### Yuu Suzumi

Thanks! That is a policy I can endorse. We want the OP to feel proud of his solution, after all.