Proof for pivot and eigenvalue sign agreement in symmetric matrices

Question

In my Linear Algebra class, we were told that for a symmetric matrix $A$, the signs of the eigenvalues are just the signs of the pivot elements. How can one prove this?

Answer 1

There are 3 kinds of diagonalization involved-given a square matrix $M$, we can look for a non-singular matrix $P$ such that (i) $P^{-1}MP$ is diagonal (diagonalizaion by similarity) (ii) such that $P$ is orthogonal and $P^TMP$ is diagonal (diagonalization by orthogonal similarity )or (iii) $P^TMP$ is diagonal (diagonalization by congruence ). A real symmetric matrix can be diagonalized by orthogonal similarity. When it is so diagonalized, the diagonal elements will be the eigenvalues. A symmetric matrix can be diagonalized by congruence over any field of characteristic $\ne 2.$ Sylvester’s inertia theorem states that for a real symmetric matrix, the number of positive terms in such a digonalization is invariant and so is the number of negative terms. Thus, since orthogonal similarity is just a special case of congruence, when we diagonalize a real symmetric matrix by congruence, the number of positive terms equals the number of positive eigenvalues and the number of negative terms equals the number of negative eigenvalues. There is a 1-1 correspondence between quadratic forms and symmetric matrices. A practical method for diagonalizing a symmetric matrix over any field of characteristic $\ne 2$ is to do one of two operations repeatedly (i) eliminate non-diagonal elements by completing the square in a variable with a non-zero diagonal term, which is a pivot operation (ii) forcing a non-zero diagonal term, when all the remaining diagonal terms are 0. This operation does not change any coefficients of variables that have already been diagonalized. Thus the positive pivots give positive diagonal terms and the negative pivots give negative diagonal terms. The correspondence is 1-1.