Question

if A,B are two matrices that verify the relationship
AB = BA -> (1)

prove that:
(AB)^n = A^nB^n

I tried:

(AB)^n = (AB)* (AB) * (AB) * (AB) * (AB) .... (AB)
from -> (1)
AB * BA * AB * BA *AB * BA.... BA
AB^2A^2B^2A^2B^2....BA
from 1
AB^2 * B^2A^2 * A^2B^ ... B = A^nB^n

Is there a math-friendly way of writing that ? or a better solution ?

Answer 1

http://prntscr.com/8r59a2

Answer 2

I don't know if this question is really worth spending too much time on it, but hey it does seem like there's gotta be a better way. So how about like this, if I show it's true for one case then we can see it'll be true for all cases:

\[A^nB^nAB=A^nAB^nB=A^{n+1}B^{n+1}\]

So I think since \(ABAB=A^2B^2\) as our base case then for all higher cases we can always step up to it with the thing I said.

Answer 3

well AB = BA so xD

Answer 4

@Empty , My professor denies using induction.

Answer 5

so its kinda pointless

Answer 6

AB AB AB AB AB
A....... BA BA BA BA BA ....B

Answer 7

Eh whatever then you're not gonna get anything better I think lol

Answer 8

@dan815 , That's what I did, but I doubt it's understood-able .
I am looking for a notation or a better solution

Answer 9

the first post u mean?

Answer 10

thsi is the same thing as kais... u can keep switchinb AB to BA and keep getting more and more A^k and B^k

Answer 11

I posted a solution associated with the question itself.

Answer 12

Yea, I understand the idea. But will I just write AB * BA * AB * BA... BA
= A^n B^n

Answer 13

no u wont do that

Answer 14

you have to show a strategy how to get more A^ks on the left and B^ks on the right

Answer 15

That's more readable, but why n+1 ?

Answer 16

A (BA ,,BA BA BA)^n B =A^1 (AB AB AB... AB)^n B^1
A (AB AB .... AB AB) B = A^2 (BA .... BA BA)^n-1 B^2

Answer 17

u lose an A and B from the middle part every step

Answer 18

this can be your induction step

Answer 19

lol you shouldn't have said that

Answer 20

Said what ?

Answer 21

Do you mean the induction word xD ?

Answer 22

show
A^k (AB)^(n+1) B^k = A^(k+1) (AB)^n B^(k+1)

Answer 23

i think its fine your teacher shouldnt have problems with this

Answer 24

Professor not teacher, He is enthusiast about giving zeros out.

Answer 25

well if u want to be a little more correct

Answer 26

u can write this expression so that there is a relationship between k and n

Answer 27

A^k (AB)^(n-k) B^k = A^(k+1) (AB)^(n-k-1) B^(k-1)

Answer 28

this way u can show one induction step and say that when k reaches n you will be left with A^n B^n

Answer 29

Well, I wish he'd like this, I hate zeros xD

Answer 30

just write this out formally

Answer 31

Show me the formal way of writing them.

Answer 32

xD

Answer 33

it*

Answer 34

that too much work

Answer 35

u get the idea, just do the best u can