proving matrix theorem

**msokol89** · March 9th 2011, 12:57 PM

Hello,
Can you please help solve the following problem:
Given two matrices A,B and AB=BA, I need to prove that for every n:
${(AB)}^{n}={A}^{n}{B}^{n}$ .

Thanks in advance,
Michael

**Plato** · March 9th 2011, 01:39 PM

Originally Posted by msokol89

Hello,
Can you please help solve the following problem:
Given two matrices A,B and AB=BA, I need to prove that for every n:
${(AB)}^{n}={A}^{n}{B}^{n}$ .

Have you tried to prove this by induction?

**topspin1617** · March 9th 2011, 05:16 PM

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

But if you just write out the definition of exponentiation

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

the result should be clear.

**msokol89** · March 9th 2011, 08:56 PM

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:
${(AB)}^{1}=(A)}^{1}(b)}^{1}$
2)I assumed for n=k it is true:
${(AB)}^{k}=(A)}^{k}(b)}^{k}$
3)I checked for n=k+1:
${(AB)}^{k+1}=(A)}^{k+1}(b)}^{k+1}$
${(AB)}\cdot {(AB)}^{k}={A}\cdot {A}^{k}\cdot {B}\cdot {B}^{k}$
Then, after switching places and using the induction assumption , I reached:
${(AB)}\cdot {(AB)}^{k}={(AB)}\cdot {(AB)}^{k}$

is this ok?

**topspin1617** · March 10th 2011, 07:55 AM

Originally Posted by msokol89

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

is this ok?

I think the last couple steps are... out of order or something.

You have

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

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

**msokol89** · March 10th 2011, 10:36 AM

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

**HallsofIvy** · March 10th 2011, 12:19 PM

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 $(AB)^{k+1}= (AB)(AB)^k= A^{k+1}B^{k+1}$ . The trivial observation that $(AB)(AB)^k= (AB)(AB)^k$ doesn't do t

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

**msokol89** · March 10th 2011, 09:59 PM

Originally Posted by HallsofIvy

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

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

ok thanks,
Do you have a way of solving the problem?

**Opalg** · March 11th 2011, 12:37 AM

Originally Posted by msokol89

Hello,
Can you please help solve the following problem:
Given two matrices A,B and AB=BA, I need to prove that for every n:
${(AB)}^{n}={A}^{n}{B}^{n}$ .

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 $AB^n = B^nA$ for every $n\geqslant1.$

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

**msokol89** · March 11th 2011, 12:51 AM

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 $AB^n = B^nA$ for every $n\geqslant1.$

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

I understand, thanks.

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