1. ## proving matrix theorem

Hello,
Given two matrices A,B and AB=BA, I need to prove that for every n:
$\displaystyle {(AB)}^{n}={A}^{n}{B}^{n}$.

Michael

2. Originally Posted by msokol89
Hello,
Given two matrices A,B and AB=BA, I need to prove that for every n:
$\displaystyle {(AB)}^{n}={A}^{n}{B}^{n}$.
Have you tried to prove this by induction?

3. I suppose to technically prove this, you would use induction.

But if you just write out the definition of exponentiation

$\displaystyle (AB)^n=ABAB\cdots AB$ ($\displaystyle n$ times)

the result should be clear.

4. Originally Posted by Plato
Have you tried to prove this by induction?
I have tried to prove it by induction:
1) I checked for n=1:
$\displaystyle {(AB)}^{1}=(A)}^{1}(b)}^{1}$
2)I assumed for n=k it is true:
$\displaystyle {(AB)}^{k}=(A)}^{k}(b)}^{k}$
3)I checked for n=k+1:
$\displaystyle {(AB)}^{k+1}=(A)}^{k+1}(b)}^{k+1}$
$\displaystyle {(AB)}\cdot {(AB)}^{k}={A}\cdot {A}^{k}\cdot {B}\cdot {B}^{k}$
Then, after switching places and using the induction assumption , I reached:
$\displaystyle {(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}$

is this ok?

5. Originally Posted by msokol89
I have tried to prove it by induction:
1) I checked for n=1:
$\displaystyle {(AB)}^{1}=(A)}^{1}(b)}^{1}$
2)I assumed for n=k it is true:
$\displaystyle {(AB)}^{k}=(A)}^{k}(b)}^{k}$
3)I checked for n=k+1:
$\displaystyle {(AB)}^{k+1}=(A)}^{k+1}(b)}^{k+1}$
$\displaystyle {(AB)}\cdot {(AB)}^{k}={A}\cdot {A}^{k}\cdot {B}\cdot {B}^{k}$
Then, after switching places and using the induction assumption , I reached:
$\displaystyle {(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}$

is this ok?
I think the last couple steps are... out of order or something.

You have

$\displaystyle (AB)^{k+1}=(AB)(AB)^k=ABA^kB^k=AA^kBB^k=A^{k+1}B^{ k+1}$.

Your last line is $\displaystyle {(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}$. But... this is clearly true, it doesn't say anything about.. anything.

6. it proves that this is true for n=k+1, therefore the entire statement is true.

7. Originally Posted by msokol89
it proves that this is true for n=k+1, therefore the entire statement is true.
No, it doesn't. You need to prove that $\displaystyle (AB)^{k+1}= (AB)(AB)^k= A^{k+1}B^{k+1}$. The trivial observation that $\displaystyle (AB)(AB)^k= (AB)(AB)^k$ doesn't do t

Having said that $\displaystyle (AB)^k= A^kB^k$, you can conclude that $\displaystyle (AB)^{k+1}= (AB)(AB)^k= (AB)(A^kB^k)$. Now, what can you conclude from that?

8. Originally Posted by HallsofIvy
No, it doesn't. You need to prove that $\displaystyle (AB)^{k+1}= (AB)(AB)^k= A^{k+1}B^{k+1}$. The trivial observation that $\displaystyle (AB)(AB)^k= (AB)(AB)^k$ doesn't do t

Having said that $\displaystyle (AB)^k= A^kB^k$, you can conclude that $\displaystyle (AB)^{k+1}= (AB)(AB)^k= (AB)(A^kB^k)$. Now, what can you conclude from that?
ok thanks,
Do you have a way of solving the problem?

9. Originally Posted by msokol89
Hello,
Given two matrices A,B and AB=BA, I need to prove that for every n:
$\displaystyle {(AB)}^{n}={A}^{n}{B}^{n}$.
It looks as though you need to use induction twice if you really want to pin this result down rigorously.

First, use induction to show that $\displaystyle AB^n = B^nA$ for every $\displaystyle n\geqslant1.$

Then use that result to prove by induction on m that $\displaystyle A^mB^n = B^nA^m$ for every $\displaystyle m\geqslant1$ (and hence in particular when m=n).

10. Originally Posted by Opalg
It looks as though you need to use induction twice if you really want to pin this result down rigorously.

First, use induction to show that $\displaystyle AB^n = B^nA$ for every $\displaystyle n\geqslant1.$

Then use that result to prove by induction on m that $\displaystyle A^mB^n = B^nA^m$ for every $\displaystyle m\geqslant1$ (and hence in particular when m=n).
I understand, thanks.