“…If, at a given grid point, we obtain A1 and a corresponding eigenvector x1, then as the Lanczos starting vector for Algorithm 3 .1 at the next grid point we take u = x1 . Some justification for this continuation strategy is given by Lui [19] . For the field of values computation, if, for a given angle 0, we obtain extremal eigenvalues A1 and a,, and corresponding eigenvectors x1 and x,,, then our starting vector for the next value of 0 is u = ( x1 + x n )/IIx1 + x,,11 2 .…”