A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Loser66

  • one year ago

Let A be a real 2 x 2 matrix If A has 2 distinct eigenvalues, then A^2 has 2 distinct eigenvalues, True of False. why? Please, help

  • This Question is Closed
  1. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @Empty

  2. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @phi

  3. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @cwrw238

  4. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I think you can write the matrix as S^-1 L S where L has the eigenvalues on the diagonal, and S is the eigenvectors A^2 = S^-1 L S S^-1 L S= S^-1 L^2 S if the values are distinct say -1 and +1 , after squaring you would get +1 +1 no longer distinct

  5. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Since it is False, we can find out a counter example to prove it False What if it is True, how to prove? Like: A is 2 x2 matrix, the determinant of A^2 is nonnegative,. How to prove?

  6. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Let the eigenvalues be -1 1 make up two eigenvectors and compute A = S^-1 L S

  7. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    example A= -3 -4 2 3

  8. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    eigs(A) ans = 1 -1 eigs(A^2) ans = 1 1

  9. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    @phi I got you. One more question, if I not lucky enough to pick the right one to manipulate, (I meant if I pick the wrong one which gives me the eigenvalues of A^2 distinct,) I fail the test, right?

  10. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    If the matrix has two eigenvalues then you can diagonalize it: \[A=PDP^{-1}\] where \[D=\left[ \begin{array}c \lambda_1 & 0\\0 & \lambda_2\\\end{array} \right]\] so you can square A to get: \[A^2= (PDP^{-1})( PDP^{-1}) = PD^2P^{-1}\] and D^2 has the form: \[D^2=\left[ \begin{array}c \lambda_1 ^2 & 0\\0 & \lambda_2^2\\\end{array} \right]\] which are the eigenvalues of D, so like @phi is saying, \[\lambda_1 = - \lambda_2 \] would be a counter example. I realizeI probably just wasted my time repeating everything that was just said because I'm a slow typer lol oh well

  11. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    In general, if the values are distinct , then the squares are distinct. So it is a special case that they become "not distinct"

  12. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    @Empty but it looks nicer

  13. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Hey!! At the end, it is T or F?

  14. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    It only takes one counter-example to make it false. The statement is false

  15. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    From @Empty we have lambda1 , lambda 2 (distinct) after then, the eigenvalues of A^2 are lambda1 = - lambda2, (distinct also) ??? I confused

  16. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    He is saying if you start with L1 = - L2 (thus distinct) for matrix A then for matrix A^2 you get L1^2 = L2^2 (no longer distinct)

  17. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Got you. Thanks a lot.

  18. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.