Typos/Clarifications in Numerical Linear Algebra by Trefethen and Bau, 2000 reprinting
p. 29, Proof of Theorem 4.1. Maximizing ||A v_1|| over ||v_1|| = 1
gives a vector whose norm is sigma_1; dividing through by sigma_1
gives us u_1, which has unit length.
p. 92, x is used on this page to mean the roots, i.e. f, while the a_i
are the data (known as x on the previous page)
p. 92, line 17. a_15 is missing a minus sign
p. 92, eqn after (12.8):
1.67*10^9 * 15^(14) /(5!*14!) = 4.6602*10^(12) NOT 5.1*10^(13)
p. 112, Ex 15.1(e). The division operator is also needed.
p. 158, middle of the page, should be "(multiplication by L2)".
p. 262 It is not explicitly stated in the text, but "lambda" is the outlier
eigenvalue.
p. 286, relationship between usual notation and authors' notation for
Legendre polynomials. A possible point of confusion is that here,
q_j+1 is a multiple of the usual Legendre polynomial P_j, but on
p.53, q_j is a multiple of P_j. The reason is that in the context
of the Lanczos 3-term recurrence, we need q_0 = 0.
p. 295 A property not mentioned in Theorem 38.1 is the obvious but crucial
fact that r_n = b - A x_n.
OTHER POSSIBLE IMPROVEMENTS
Lecture 6:
2nd half of proof of Theorem 6.1
since X*P*(I-P)Y=0 (here * does not mean times, it's the adjoint operation.)
therefore P*(I-P) must be zero(this is obvious and very easily to prove),
which directly leads to the conclusion P=P* (also very obvious because here
we have P*=P*P, so P=P*P also) (from Lin Shi)
Lecture 24:
Could have defined minimal polynomial along with the characteristic polynomial.
Is it defined anywhere before it is used in Lectures 33 and 34?
Lecture 38:
Why does no one ever mention that CG is a misnomer? The gradients
(of 1/2 x'Ax - x'b, i.e. the negative residuals -r_n = Ax_n - b),
are ORTHOGONAL, not conjugate. It's the p_n that are A-conjugate.
There is a beautiful connection between CG and Lanczos that is not
mentioned in the text. Let x in K_n, writing x = Q_n y, where the columns
of Q_n are the Lanczos vectors for A. Since x_n solves
min 1/2 x'Ax - x'b over x in K_n,
the corresponding y must solve the problem
min 1/2 y' Q_n' A Q_n y - b'Q_n y over all y
and, since Q_n' A Q_n = T, which is tridiagonal and also positive definite,
the minimizer is given by
T y = Q_n' b (= [1 0 ... 0]' if b = q_1)
The Cholesky factorization of T is LL', where L is lower triangular and
bidiagonal, so we need to solve
LL' y = [1 0 ... 0]'
Doing so gives an equivalent, not-so-standard version of CG. See Golub and
Van Loan for details. This idea is due to Paige and Saunders.