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Matrices Questions

  1. Sep 26, 2011 #1
    Is any symmetric matrix diagonalisable with an orthogonal change of basis?

    Does the minimal polynomial of any real matrix split into distinct linear factors?

    Is a real inner product an example of a bilinear form?

    Could 2 complex matrices which are similar have different Jordan normal forms?
     
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  3. Sep 26, 2011 #2

    micromass

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    So, what did you try already??
     
  4. Sep 26, 2011 #3
    I just want to know 'yes' or 'no' out of interest
     
  5. Sep 26, 2011 #4

    micromass

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    For the first one, let A be symmetric. Let u be an eigenvector with eigenvalue [itex]\lambda[/itex] and let w be an eigenvector with eigenvalue [itex]\mu[/itex]. Can you calculate

    [tex]<Au,w>[/tex]

    in several ways?
     
  6. Sep 26, 2011 #5
    I think YES, NO, YES, NO respectively. Is that correct?
     
  7. Sep 26, 2011 #6

    micromass

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    Why do you think that?
     
  8. Sep 26, 2011 #7
    I think the 4th question is worded slightly ambiguously; it's probably safe to assume that it means "different up to reordering of the Jordan blocks" (in which case you're correct). But if it doesn't, then [tex]\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}[/tex] and [tex]\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}[/tex] would be an example of similar matrices with different JNFs.

    I know the first 3 are right - yes, no, yes but for the 4th one, do you think that the answer to the last one is no?
     
  9. Sep 26, 2011 #8

    micromass

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    If you mean "different up to reordering of the Jordan blocks" then you are correct. Similar matrices have the same Jordan canonical form then.

    This can be seen by putting [itex]A=S^{-1}BS[/itex]. If C is the Jordan basis, then

    [tex]CAC^{-1}[/tex]

    is the Jordan canonical form of A. Now, can you find easily the Jordan canonical form of B (just plug the expressions into each other).
     
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