# Is there a quick way to determine the eigenvalues of a symmetric matrix?

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?

• 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. – James S. Cook Mar 30 '17 at 2:39
• Also of help: the eigenvalues of a real symmetric matrix are all real valued. – erfink Mar 30 '17 at 2:42
• @JamesS.Cook yup that's true. However, how about the other non 0 eigenvalues? – dzl Mar 30 '17 at 2:48
• This question is very similar – Mark Mar 30 '17 at 2:50