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.