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 agoim new plz tell me what does ^ means

8 years agoIt means "to the exponent of" so whatever comes before that symbol is raised to the exponent of whatever comes after that symbol

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

8 years agoNow I'm just working on finding B^k and A^k

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

8 years agoI mean the "proof" of the formula.....

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

8 years agoAnd \[A ^{k}=P ^{-1}B ^{K}P\] 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 agoSorry, I've been having internet issues....

8 years agoNo problem, I am too

8 years agoYou'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 agookay Ill just do \[B ^{k}=BB ^{K-1}B ^{K-2}...B ^{K-K}\]

8 years agoFor 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 agoI'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 agoThanks for everything i'm good now

8 years ago