# Similar Matrices to the k power

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

p(k):= Bk=S-1*Ak*S
p(k+1):= Bk+1=S-1*Ak+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.

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

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

You don't need induction here.
Answer the previous poster's question, what is Bk 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.

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

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!

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

Write it out.

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.

Now try it for (P-1AP)2=.....

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

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

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.

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.

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

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?

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

This is the step I am stuck on.

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-1AkP)(P-1AP) = P-1Ak(PP-1)AP = ......