Similarity is transitive

  • Thread starter FourierX
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  • #1
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Homework Statement



In linear algebra, how to prove that the Similarity is transitive?



Homework Equations



[T]C = P-1 [T]B P

The Attempt at a Solution


My logic: if a is similar to b and b is similar to c, then a is similar to c. But how do you implement that in case of similarity ?
 
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Answers and Replies

  • #2
Dick
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If A is similar to B, A=PBP^(-1). If B is similar to C, B=QCQ^(-1). Where P and Q are invertible. Put the second equation into the first equation.
 
  • #3
ptr
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If A = P B Pinv and B = K C Kinv, where Pinv is the inverse of P, and mutatis mutandis for K, use the property that (PK)inv is (Kinv)(Pinv). and hence A = (PK) C (PK)inv.
 
  • #4
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how do PQ and P-1Q-1 behave ?
 
  • #5
Dick
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You should have asked how do PQ and Q^(-1)P^(-1) behave. Whatever 'behave' means. They don't do anything, they are just invertible matrices. And one is the inverse of the other, see ptr's post.
 

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