0
$\begingroup$

In a problem set, we were asked to determine the eigenvalues of a $5 \times 5$ symmetric matrix. I know its possible to just find the eigenvalue, $ \lambda $ by calculating the values of $\lambda $ which satisfy $\det( \lambda I - A ) = 0 $, where A is the aforementioned $ 5 \times 5$ matrix.

However, is there a quicker and easier way to find the eigenvalues of a symmetric matrix?

$\endgroup$
  • 1
    $\begingroup$ trace is the sum of the eigenvalues and the determinant is the product of the eigenvalues. Sometimes helpful. If you have a repeated row or column or any sort of linear dependence amongst the columns or rows then the determinant is zero hence there is an eigenvalue of zero. $\endgroup$ – James S. Cook Mar 30 '17 at 2:39
  • 1
    $\begingroup$ Also of help: the eigenvalues of a real symmetric matrix are all real valued. $\endgroup$ – erfink Mar 30 '17 at 2:42
  • $\begingroup$ @JamesS.Cook yup that's true. However, how about the other non 0 eigenvalues? $\endgroup$ – dzl Mar 30 '17 at 2:48
  • 1
    $\begingroup$ This question is very similar $\endgroup$ – Mark Mar 30 '17 at 2:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.