1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Non-degenerate Hermitian Matrices and their Eigenvalues

  1. Nov 29, 2015 #1

    RJLiberator

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    Is there a non-degenerate 2x2 matrix that has only real eigenvalues but is not Hermitian? (Either find such a matrix, or prove that it doesn't exist)

    2. Relevant equations


    3. The attempt at a solution

    Here's my problem. I'm getting Contradicting results.

    So, I found this 2x2 matrix:
    \begin{matrix}
    1 & 2 \\
    3 & 2
    \end{matrix}

    This matrix has eigenvalues of 4 and -1.

    This matrix is not hermitian as the hermitian representation of this matrix transposes it.

    Therefore this matrix is a 2x2 matrix that is non-degenerate and is NOT hermitian and has ONLY real eigenvalues.

    This should be the example that the question was looking for.

    However, in my notes, I have the equation for eigenvalues being
    eigenvalue = [(a+b)+/- (squareroot ((a-b)^2-4cd))]/2

    where
    \begin{matrix}
    a & c \\
    d & b
    \end{matrix}

    And if we take this representation, we get negative eigenvalues.

    What is going on here?
     
  2. jcsd
  3. Nov 30, 2015 #2

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    Are you sure about that equation?
    I find, by expanding ##\det(A-\lambda I)=0## and solving the quadratic equation:
    ##\lambda=\frac{(a+b)\pm\sqrt{(a+b)²-4(ab-cd)}}{2}##
     
  4. Nov 30, 2015 #3

    RJLiberator

    User Avatar
    Gold Member

    Yes, I agree with what you are saying, and I noticed that as well, however, if you simplify it further:

    (a+b)^2 = a^2+b^2+2ab
    So then, 2ab-4ab becomes -2ab.

    so now we have a^2+b^2-2ab-4cd
    but a^2+b^2-2ab is just (a-b)^2 and so we get
    (a-b)^2-4cd.

    which is why I am getting contradictory results.
    What is going wrong?
     
  5. Nov 30, 2015 #4

    Samy_A

    User Avatar
    Science Advisor
    Homework Helper

    It should be ##+4cd## after the simplification.
     
  6. Nov 30, 2015 #5

    RJLiberator

    User Avatar
    Gold Member

    Ah....yes, that was the source of my frustration.

    So, it seems my example now holds up and has nothing contradicting it.

    Thank you kindly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Non-degenerate Hermitian Matrices and their Eigenvalues
  1. Degenerate Eigenvalues (Replies: 7)

  2. Hermitian matrices (Replies: 2)

  3. Non-Hermitian matrices (Replies: 1)

Loading...