(11/16/98) Let A be an matrix. Show that , where denotes the transpose of A.

Solution:

First note that since is an matrix, the determinant is defined. Next, note that is symmetric since . By the Spectral Theorem for Symmetric Matrices, is orthogonally diagonalizable; that is, there exist an orthogonal matrix P and a diagonal matrix D (both ) such that . The columns of P form an orthonormal basis of made up of eigenvectors of , with the diagonal entries of D as the corresponding eigenvalues. Thus, if is the ith column of P, then

where is the ith diagonal entry of D. That means the diagonal entries of D are all nonnegative, and .