Prove the following claim: If A is a symmetric nn matrix, and all of A's eigenvalues are strictly positive, then ^ > 0, for all nonzero vectors ^ .

Hints:

1. If A is symmetric, then A = A^T (by definition), and A will be "diagonalizable"

2. The first step in the proof will be to show that A's symmetry ensures that the inverse of A's eigenvector matrix, E, is the transpose of A's eigenvector matrix. (i.e. - that A = A^T E^1 = E^T )