Proving Similarity of Matrices: A^2=C to B^2=C

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Homework Help Overview

The discussion revolves around proving the similarity of matrices, specifically showing that if \( A^2 = C \) for an \( n \times n \) matrix \( A \) and \( B \) is similar to \( A \), then \( B^2 \) must also equal \( C \). Participants are exploring the implications of different values for \( C \), including the zero matrix and multiples of the identity matrix.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants discuss specific examples of matrices \( A \) and \( B \) to challenge the validity of the statement. Questions arise about the nature of \( C \) and whether it could be the zero matrix or a multiple of the identity matrix, which could affect the proof.

Discussion Status

The discussion is active, with participants questioning the assumptions in the original problem statement. Some have provided examples that suggest the statement may not hold true under certain conditions, while others are exploring how different definitions of \( C \) could change the outcome.

Contextual Notes

There is uncertainty regarding whether \( C \) is meant to represent the zero matrix or a different form, as indicated by the slant of the letter 'O' in the original text. This ambiguity is influencing the direction of the discussion.

eyehategod
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I need help with this proof. Can anyone lead me in the right direction?

Let A be an nxn matrix such that A^2=C.
Prove that if B~A, then B^2=C.
 
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Let A=diag(1,2), B=diag(2,1). A^2=diag(1,4), B^2=diag(4,1). A~B. A^2 is not equal to B^2. Is there something you aren't telling us about C or do you want to prove A^2~B^2?
 
I gave you exactly what the book says
 
eyehategod said:
I gave you exactly what the book says

If that's exactly what the book says, then you can't prove it. Because it's false.
 
what if C were to be 0. Would that proof make sense?
 
eyehategod said:
what if C were to be 0. Would that proof make sense?

It makes all of the difference in the world. If A^2=0 and A~B then B^2=(PAP^(-1))(PAP^(-1)). What's that?
 
that would be:
B^2=P^(-1)A^(2)P
B^2=P^(1)0P=0
 
It would also be very different if C were any other multiple of the identity matrix. Wouldn't it?
 
THe book really has instead of A^2=C its A^2=O.But I can't tell if its zero or the letter O. THe O is at a slant if that means anything
 
  • #10
eyehategod said:
that would be:
B^2=P^(-1)A^(2)P
B^2=P^(1)0P=0

Yes.
 
  • #11
eyehategod said:
THe book really has instead of A^2=C its A^2=O.But I can't tell if its zero of the letter O. THe O is at a slant if that means anything

I don't know. But as I said any multiple of the identity would work as well as 0.
 

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