Is there a way to show that Ak is unitarily similar to Bk using induction?

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Homework Help Overview

The discussion revolves around the concept of unitary similarity between matrices, specifically focusing on proving that if matrix A is unitarily similar to matrix B, then their powers Ak and Bk are also unitarily similar for all k. The participants are exploring the implications of this relationship using induction.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using induction to establish the similarity for powers of matrices, with initial attempts focusing on the base case and the inductive step. Questions arise regarding the correct application of the unitary transformation and the manipulation of matrix products.

Discussion Status

The discussion is ongoing, with participants providing insights and corrections to each other's reasoning. Some guidance has been offered regarding the structure of the proof, but there is no explicit consensus on the next steps or final form of the proof yet.

Contextual Notes

There is an emphasis on the need to correctly apply the properties of unitary matrices and their adjoints, as well as the challenge of maintaining clarity in the inductive argument. Participants are also navigating the constraints of the problem as a homework assignment.

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Homework Statement


SUppose A and B are nxn matrices in the complex field and that A is unitarily similar to B.

Homework Equations


Show that Ak is unitarily similar to Bk for all k=1,2,3,..


The Attempt at a Solution


I used induction to show its true for k=1 which it is.
Then for k=n+1,
An+1=(U*)n+1Bn+1Un+1
AnA=(Un)*(U*)B(Bn)UnU.

That is as far as i got, any help?
 
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You want A^k = U^\dagger B^k U. You don't need to take U and its adjoint to the k-th power.
 
so then would it be simply
An+1=U*Bn+1U?? and I am finished?
 
Well, you need to prove that statement is true.
 
so i would have
AnA=U*BnBU
 
i am stuck??=[
 
chuy52506 said:
so i would have
AnA=U*BnBU
Nope, you have, using the induction hypothesis An=U*BnU,

An+1 = AnA = (U*BnU)A

Now write that last factor of A in terms of U and B.
 

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