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Hero

If a 2x2 matrix has the same two eigenvalues as another 2x2 matrix, what conclusions can you infer about the two matrices? Make some hypotheses and attempt to prove them, or refute them via counterexamples.

  • one year ago
  • one year ago

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  1. malevolence19
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    \[\left[\begin{matrix}\psi & \xi \\ \chi & \zeta\end{matrix}\right] \implies (\psi - \lambda)(\zeta - \lambda) - \xi * \chi = 0\] Take the transpose of that matrix you have: \[\left[\begin{matrix}\psi & \xi \\ \chi & \zeta\end{matrix}\right]^T=\left[\begin{matrix}\psi & \chi \\ \xi & \zeta\end{matrix}\right] \implies (\psi - \lambda)(\zeta - \lambda) - \xi * \chi = 0\] They are the same. I'm not sure if you can ALWAYS assume this but it seems pretty logical. I also don't know any theorems to name.

    • one year ago
  2. Hero
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    Brilliantly done!

    • one year ago
  3. eliassaab
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    No, they are not the same always.

    • one year ago
  4. Hero
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    Hmmm

    • one year ago
  5. Hero
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    I need a counterexample.

    • one year ago
  6. Hero
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    Well, by default they wouldn't be the same matrices

    • one year ago
  7. eliassaab
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    \left( \begin{array}{cc} 2 & 4 \\ 0 & 3 \\ \end{array} \right) \\ \left( \begin{array}{cc} 2 & 0 \\ 0 &3 \\ \end{array} \right) Have the same eigenvalues 2 and 3 but they are not the same.

    • one year ago
  8. Hero
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    The matrices that malevolence posted are not the same either are they?

    • one year ago
  9. eliassaab
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    Ok. I thought he wanted to conclude that if two matrices have the same eigenvalues then they are the same. This is not true in general. What do you think the answer to your question is?

    • one year ago
  10. Hero
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    Well, apparently, there's more than one conclusion. If you have two matrices that are not the same but have the same eigenvalues, I'm not really sure what to conclude which is why I posted the question.

    • one year ago
  11. eliassaab
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    For example they cannot be similar either. Can you construct a counter example?

    • one year ago
  12. Zarkon
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    same rank

    • one year ago
  13. eliassaab
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    No, you cannot conclude that they have the same rank

    • one year ago
  14. eliassaab
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    \[ \left( \begin{array}{cc} 0 & 0 \\ 0 & 0 \\ \end{array} \right)\\ and \\ \left( \begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array} \right) \] have the same eigenvalue but are not similar.

    • one year ago
  15. Hero
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    So what exactly am I supposed to conclude?

    • one year ago
  16. Zarkon
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    that they have the same eigenvalues ;)

    • one year ago
  17. eliassaab
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    The have the same determinant. They have the same characteristic polynomial. Something like that.

    • one year ago
  18. Hero
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    So any two matrices with the same eigenvalues will have the same determinant and characteristic polynomial? That doesn't seem accurate. Maybe for certain cases, but not for every case

    • one year ago
  19. Zarkon
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    the product of the eigenvalues is always equal to the determinalt for any nxn matrix

    • one year ago
  20. eliassaab
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    @Zakron answered you correctly.

    • one year ago
  21. Zarkon
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    the trace will be the same

    • one year ago
  22. eliassaab
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    The trace is the sum of eigenvalues.

    • one year ago
  23. Zarkon
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    the trace is the sum of the main diagonal

    • one year ago
  24. eliassaab
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    but it is equal to the sum of the eigenvalues.

    • one year ago
  25. Zarkon
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    by theorem..not my definition

    • one year ago
  26. eliassaab
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    Of course.

    • one year ago
  27. Zarkon
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    that is what I was pointing out

    • one year ago
  28. eliassaab
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    I see/

    • one year ago
  29. eliassaab
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    @hero are you satisfied with what is posted above.

    • one year ago
  30. Hero
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    yes,thanks

    • one year ago
  31. eliassaab
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    yw

    • one year ago
  32. Hero
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    I can only award one medal

    • one year ago
  33. Zarkon
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    I think that if you are green (or purple) you should be able to give more than one medal.

    • one year ago
  34. Hero
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    I agree

    • one year ago
  35. TuringTest
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    I don't think I can elaborate much on what Zarkon said do you have a specific question?

    • one year ago
  36. Hero
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    The question I had for you Turing had nothing to do with this question.

    • one year ago
  37. malevolence19
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    I love this conversation haha @TuringTest @Zarkon @eliassaab

    • one year ago
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