Class 31



  1. Two tricks to show that:
    1. A is m by n, B is n by m; m <= n; then the eigenvalues of AB and BA are the same but with BA having n-m extra zero eigenvalues.
    2. Suppose A is n by n and t is an eigenvalue of A. Then the nonzero columns of the adjoint adj (tI-A) are eigenvectors of A for the eigenvalue t.
  2. diagonalization of the matrix of a quadratic form x'Ax, A=A' (also: A=A' without loss of generality)