Similar Matrices to the k power

[Dustinsfl]

If A and B are similar matrices, show that A^k and B^k are similar.
I am almost positive this has to be done by induction.

p(k):= B^k=S^-1*A^k*S
p(k+1):= B^k+1=S^-1*A^k+1*S
Assume p(k) is true.

I know I could take p(k) and multiply right by A but I don't think that will go any where.

 

[Tinyboss]

If $$B=P^{-1}AP$$, then $$B^k=$$... ?

 

[Dustinsfl]

I forgot to mention I am proving this for each positive integer or the natural numbers.

 

[VeeEight]

You don't need induction here.
Answer the previous poster's question, what is B^k in terms of A?
If you can't get that right away, try it for a smaller integer. If B=P^-1AP, what is B²? You should see how the answer follows.

 

[Dustinsfl]

So I am going to obtain s inverse to the k a to the k and s to the k which is equal to b to the k

 

[Tinyboss]

So I am going to obtain s inverse to the k a to the k and s to the k which is equal to b to the k

No, matrices don't commute under multiplication in general.

Edit: Although, if they did, you'd be home free, too...but they don't!

 

[Dustinsfl]

What should I do then with (S^-1*A*S)^k then?

 

[Tinyboss]

Write it out.

 

[VeeEight]

Try small powers. For example, k=2
(XYZ)²=(XYZ)(XYZ)=XYZXYZ
Now try it for (P^-1AP)²=...

After you have that done you can generalize to the power of k.

 

[Dustinsfl]

Now try it for (P^-1AP)²=...

I not sure about this part.

 

[VeeEight]

Just multiply the term by itself.
(XYZ)²=(XYZ)(XYZ)=XYZXYZ

 

[Dustinsfl]

Never mind my eyes were playing tricks on me I thought that was p- not to the power.

 

[Dustinsfl]

I see how it works since the middle terms become I which goes away. I now that this will happen for where each last term meets the first time but I am still unsure how to prove it.

 

[VeeEight]

If you see it for the smaller powers (for k=2,3.. etc), it is easy to generalize to the power of k. You can use induction if you want but it is not really needed.

 

[Tinyboss]

You will use induction after all. But now that you see the pattern, you should be able to make it work.

 

[Dustinsfl]

If I use my term to the k+1, what do I do after I separate it to term to the k times term to the first?

 

[VeeEight]

(P^-1AP)^k+1 = (P^-1AP)^k(P^-1AP). Then apply the induction hypothesis (assuming you proved a base case).

 

[Dustinsfl]

This is the step I am stuck on.

 

[VeeEight]

What exactly are you stuck on? Apply the induction hypothesis to the term (P^-1AP)^k. This will give you the familiar paring of the P and P^-1 terms when you multiply it by P^-1AP.

So you have (P^-1AP)^k(P^-1AP) = (P^-1A^kP)(P^-1AP) = P^-1A^k(PP^-1)AP = ...
Your result follows from this.

 

[Mark44]

By assumption, B^k = P^-1A^kP.

So B^k+1 = B^kB = ?

 

[Dustinsfl]

I was working the wrong side of the equation was the issue thanks.