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Homework Help: Jordan canonical form

  1. Jul 31, 2008 #1
    1. Show that two matrices A,B ∈ Mn(C) are similar if and only if they share a Jordan canonical form.

    2. Prove or disprove: A square matrix A ∈ Mn (F) is similar to its transpose AT. If the statement is false, find a condition which makes it true.
    (I'm pretty sure that this is true and can be proven using the above by showing that A and AT share a Jordan form.)

    I have a basic understanding of what Jordan blocks are and what JCF matrices look like, but I don't know what the identifying characteristics of a specific JCF are or how to show that two arbitrary matrices share the same form. I know in both questions that the two matrices have the same characteristic polynomial and spectrum, but I don't know where to go from there.
     
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  3. Aug 1, 2008 #2

    CompuChip

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    So lets start with the direct implication. Suppose that two matrices A and B share the same JNF. What does this mean explicitly? If A' is a JNF of A, then there exists a matrix P such that [itex]A' = P^{-1} A P[/itex] and A' has a specific form, right? So start working this out and try to show that they are similar (what does it mean explicitly if two matrices are similar?)
     
  4. Aug 1, 2008 #3
    But the fact that there exists a matrix P such that A'= P-1AP, is the direct definition of similarity and the proof would be trivial. Does it follow directly from the definition of JNF that such a matrix P always exists?
     
  5. Aug 1, 2008 #4

    CompuChip

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    No, it's not trivial (very easy, though). If, in what you wrote down, A' is the Jordan normal form, such a matrix P exists and indeed you express there that A is similar to its JNF. So you can assume that. Similarly, suppose that there is a(n invertible) matrix Q such that Q-1ATQ is the same Jordan normal form (and Q is the similarity transformation between the transpose of A and that normal form). Now you still have to prove that A and AT are similar, that is: construct a matrix S such that A = S-1 AT S.
    In fact this is a more general statement: if A is similar to C, and B is similar to C, then A is similar to B. You can try and prove that instead (it's actually the same proof as you don't have to use that C is in JNF anywhere). (In fact I think that similarity is even an equivalence relation, and the direct implication of your original question also shows the "hardest" part of that :smile:)
     
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