Mathematics OpenStudy (anonymous):

Hey everyone..P, A, and B are nXn matrices and have the equation B=P^-1AP (so they are similar) Show that B^2=P^-1A^2P and find B^k and A^k

8 years ago OpenStudy (anonymous):

im new plz tell me what does ^ means

8 years ago OpenStudy (anonymous):

It means "to the exponent of" so whatever comes before that symbol is raised to the exponent of whatever comes after that symbol

8 years ago OpenStudy (anonymous):

Couldn't you just write, \[ B^{2} = (P^{-1} A P)(P^{-1} A P) \] and then use the fact that P and \(P^{-1}\) are inverses to write, \[ B^{2} = P^{-1} A (PP^{-1}) A P = P^{-1} A ( I ) A P = P^{-1} A^{2} P \] A similar arguement should get you a formula for \( B^{k} \). Once you have the formula for \( B^{k} \) you should be able to multiply that by P on the left and \( P^{-1} \) on the right you should get a formula for \( A^{k} \). Or at least you will if my quick thoughts on \( B^{k} \) are correct.

8 years ago OpenStudy (anonymous):

Thanks a lot! Yes whatever B is raised to, A is raised to as well and the P's stay the same.

8 years ago OpenStudy (anonymous):

Now I'm just working on finding B^k and A^k

8 years ago OpenStudy (anonymous):

I haven't thought much more about it, but the formula for \( B^{k} \) should be just an extention of the \( B^{2} \).

8 years ago OpenStudy (anonymous):

I mean the "proof" of the formula.....

8 years ago OpenStudy (anonymous):

sorry just I haven't been great with proofs, I've got \[B ^{k}= B B ^{K-1} which -> =(P ^{-1}AP)(P ^{-1}AP)^{K-1}\] Then \[-> =(P ^{-1}A ^{2}P)^{K-1}\]...Is that sufficient proof? I don't know where to end it

8 years ago OpenStudy (anonymous):

.....Sorry I think the answer is just \[B ^{k}=P^{-1}A ^{K}P\]...and that is sufficient

8 years ago OpenStudy (anonymous):

And \[A ^{k}=P ^{-1}B ^{K}P\] correct? I'm not sure if we need more proof than that, it just says "find an analogous relationship involving B^k and A^k

8 years ago OpenStudy (anonymous):

Sorry, I've been having internet issues....

8 years ago OpenStudy (anonymous):

No problem, I am too

8 years ago OpenStudy (anonymous):

You've got the correct formula for \(B^{k}\). I think the "best" proof (although that is relative) is just to do something like, \[ B^{k} = B B .... B B \] where you multiply the B k times then just replace each B with \( P^{-1} A P \) and cancel each \( P P^{-1} \) as above. The problem is more one of writing this out. It's more of a "thought" proof I suppose. It's probably going to depend on what you actually need to do. If It's just find a formula then you probably don't need a lot of proofs...

8 years ago OpenStudy (anonymous):

okay Ill just do \[B ^{k}=BB ^{K-1}B ^{K-2}...B ^{K-K}\]

8 years ago OpenStudy (anonymous):

For the \(A^{k}\) I think you've got the P's backwards. Remember that you've got to be careful with "order" of multiplication. I.e. you need to multiply on the same side of the equation. So, starting with the formula for \(B^{k}\) we can do the following \[ B^{k} = P^{-1} A^{k} P \] The multiply the left side by P and the right side by \(P^{-1}\). \[ P B^{k} P^{-1} = P P^{-1} A^{k} P P^{-1} \] which gives, \[ P B^{k} P^{-1} = A^{k} \]

8 years ago OpenStudy (anonymous):

I'm going to be away from my computer for a while now unfortunately. I'll try to check back later, but I don't knwo when I'll get the chance....

8 years ago OpenStudy (anonymous):

Thanks for everything i'm good now

8 years ago
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