Eigenvalues of symmetric matix

Question

Well for a symmetric matrix ($A^T = A$), is there an easy algorithm to get the eigenvalues by hand?

Especially for moderate sized matrices like 6*6 (where calculating/solving the determinant becomes infeasible).

Answer 1

Sorry - the answer below is not the answer you are looking for - in my haste I confused congruence and similarity for a moment there, but anyway - maybe I should leave the answer up and someone else can maybe add to it - especially in terms of the relationship between diagonal congruence and diagonal similarity, if there is any more to be said.

My answer below is regarding congruence and does not yield the eigenvalues as requested by OP. I know for real symmetric matrices, through Sylvester's law we can then determine the number of positive and negative eigenvalues, but no way to compute them directly from the matrix $D$ below?

There is a method where you employ only elementary column and row operations to find a diagonal matrix $D$ which is congruent to $A$, but unfortunately this is not the same as the matrix containing the eigenvalues. This method will also supply you with the matrix $Q$ so that $Q^TAQ=D$ - the source that I am consulting is Linear Algebra, fourth edition by Friedberg et al, p.431-433...here is the method - I will just do it on a simple $2 \times 2$ matrix, say we have: \begin{equation} \begin{bmatrix} 2 & 3 \\ 3 & 4 \end{bmatrix}, \end{equation}construct the augmented matrix: \begin{equation} \begin{bmatrix} 2 & 3 & | & 1 & 0\\ 3 & 4 & | & 0 & 1 \end{bmatrix}. \end{equation} Now perform an elementary column operation with the goal to reduce entry $(1,2)$ to $0$: \begin{equation} \begin{bmatrix} 2 & 0 & | & 1 & 0\\ 3 & -1/2 & | & 0 & 1 \end{bmatrix}. \end{equation} Now perform an equivalent elementary row operation: \begin{equation} \begin{bmatrix} 2 & 0 & | & 1 & 0\\ 0 & -1/2 & | & -3/2 & 1 \end{bmatrix}. \end{equation} Now the matrix on the left is the diagonal matrix congruent to the symmetric matrix we started with and the matrix on the right is $Q^T$