(*)  Suppose that     B = T ^-1AT    and that   x   is an eigenvector of   A   with corresponding
eigenvalue   \lambda -- this means that  Ax = \lambda x;   A  is an    n x n    matrix,
x   is an n-dimensional column vector and   \lambda    is a real (or complex) number.
Show that   T ^-1x   is an eigenvector of   B   with the same corresponding eigenvalue   \lambda.
This helps explain why, in Long Homework #1, we defined eigenvalues of matrices in terms
of similar diagonal matrices.  It turns out to be equivalent to the definition in the book.