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Trace, determinant, and eigenvalues 3x3

  1. Apr 25, 2010 #1
    Use the trace and determinant to compute eigenvalues.

    I know how to do this with a 2x2 but not sure how to do it with a matrix of nxn where n>2.

    \frac{1}{2} & \frac{1}{3} & \frac{1}{5}\\
    \frac{1}{4} & \frac{1}{3} & \frac{2}{5}\\
    \frac{1}{4} & \frac{1}{3} & \frac{2}{5}
    \end{bmatrix}[/tex] the det=0 and the trace=[tex]\frac{37}{30}[/tex]
  2. jcsd
  3. Apr 25, 2010 #2
    I don't think it's possible for n>2 cases, you can probably work out infinite number of matrices with the same trace & det, but with different eigenvalues
  4. Apr 25, 2010 #3
    You can't use only the determinant and trace to find the eigenvalues of a 3x3 matrix the way you can with a 2x2 matrix. For example, suppose that det(A) = 0 and tr(A) = t. Then any matrix of the form:

    a & 0 & 0\\
    0 & t-a & 0\\
    0 & 0 & 0\\

    has trace = t and determinant 0 with eigenvalues a and t-a. So you'll have to go back to the matrix to find the eigenvalues.
  5. Apr 25, 2010 #4
    According to the book it can be done with this matrix.
  6. Apr 25, 2010 #5
    Epiphany: since the determinant is 0, one of the eigenvalues has to be 0.
  7. Apr 25, 2010 #6
    Now if we could just divide the determinant of the matrix by zero and get back [tex]\lambda_1 \lambda_2 [/tex]!
  8. Apr 26, 2010 #7
    Then I got no clue, sorry.
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