# Matrix exponent rules

1. Mar 20, 2014

### negation

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

Is it true that:

(ABA^-1)^8 = AB^8A^-1 for all n x n matrix and not just for invertible matrix?

My attempt:

(ABA^-1)^8 =A^8B^8A^-8

2. Mar 20, 2014

### Staff: Mentor

That's not correct. You can't "distribute" the exponent like that for matrices.

Start with a simpler problem: what is $(A B A^{-1})^{2}$?

3. Mar 20, 2014

### negation

I can't see what is written. But I presume you wrote (ABA^-1)^2 based on my intuition of the code.

(ABA^-1)^2 = ((AB)A^-1)^2 = (AB)^2 (A^-1)^2 = (B^2A^2)A^-2

4. Mar 20, 2014

### Staff: Mentor

You'll have to find a way to solve that problem, otherwise you'll have difficulty reading most posts in PF!

Again, you can't distribute the exponent like that, because multiplication of matrices is not commutative.

Go back to first principles: A2 = AA

5. Mar 20, 2014

### negation

I haven't been able to troubleshoot it because it occurs at certain chances on the same computer. But usually it works fine.

((AB)A^-1)^2 = (ABA^-1) (ABA^-1)

6. Mar 20, 2014

### Staff: Mentor

Good. Now simplify that expression.

7. Mar 20, 2014

### negation

AABBA^-1A^-1

Is this a valid operation?

8. Mar 20, 2014

### Staff: Mentor

Absolutely not! Again, matrix multiplication is not commutative: AB ≠ BA. You can't move around the different matrices, like you would do for ordinary numbers.

What you have is

(ABA-1)(ABA-1) = A B A-1 A B A-1

Is there anything in there you can simplify?

9. Mar 20, 2014

### negation

aa^-1a^-1abb

Should I try to get it simplified to identity matrix?

10. Mar 20, 2014

### Staff: Mentor

Please, you are not allowed to move things around. You have to take every matrix in the order it appears.

What I want you do to is to find two consecutive matrices that you can simplify into something else.

11. Mar 20, 2014

### negation

Am I allowed to introduce matrices?

12. Mar 20, 2014

### Staff: Mentor

Yes, if they do not change the result, either because you do the same on both sides of an equality (meaning you can only multiply on the extrement left or extreme right), or by introducing the identity matrix I (which is neutral in terms of multiplication).

But, in your case, you do not need to do that. What do you find in the middle of the product?

13. Mar 20, 2014

### negation

There's A^-1A which can be simplified to an identity I.

If this is valid, I have ABBA^-1, and base on def of first principle

BB = B^2

this gives

AB^2A^-1

14. Mar 20, 2014

### negation

AB^2A^-1 is correct. Very positive.
Would it also be valid to say that (ABC)^-1 =/= (A^-1B^-1C^-1)?

If the exponent is more than 2, is there an abstract proof I can perform without having to flood the paper with Ai=1 .. . . .. . .Ai=n?

15. Mar 20, 2014

### Staff: Mentor

For an exponent 2n, you have
(A B A-1)2n = ((A B A-1)2)2n-1 = (A B2 A-1)2n-1 which you can use to start the proof.

You can also show the result for an exponent 2, then 4, then get the general formula by induction.

16. Mar 20, 2014

### negation

On a related note,

Would it also be valid to say that (ABC)^-1 == (A^-1B^-1C^-1)?
Since operation between inverse are different?

17. Mar 20, 2014

### Staff: Mentor

You have to reverse the order of the matrices:

(AB)-1 = B-1A-1

18. Mar 20, 2014

### negation

Yes I understand I have to. The conudrum comes in when there are more than 2 matrix.

But I'd give it a try:

(ABC)^-1 = C^-1 B^-1 A^-1

19. Mar 20, 2014

### Staff: Mentor

Exactly. Just take them two at a time:

(ABC)-1 = ((AB)C)-1= C-1(AB)-1 = C-1B-1A-1

20. Mar 20, 2014

### negation

Just bear with me for a moment.

A^-1 = BC then C^-1 = AB is true.

(A^-1)^-1 = (BC)^-1 = C^-1 B^-1

(Inverse of an inverse is the original matrix)

A = C^-1 B^-1

I'm a little confused here. I know I have to multiply both sides but B but nebulous as regard the order.

But from what I at least know:

AB = C^-1 B^-1 B

AB = C^-1