Solutions for Chapter 4: Exponential and Logarithmic Functions

Independent vectors VI, ... , v d whose linear combinations give each vector in V as v = CIVI + ... + CdVd. V has many bases, each basis gives unique c's. A vector space has many bases!

Put CI, ... ,Cn in row n and put n - 1 ones just above the main diagonal. Then det(A - AI) = ±(CI + c2A + C3A 2 + .•. + cnA n-l - An).

dij = 0 if i #- j. Block-diagonal: zero outside square blocks Du.

A = S-1 AS. A = eigenvalue matrix and S = eigenvector matrix of A. A must have n independent eigenvectors to make S invertible. All Ak = SA k S-I.

The first nonzero entry (the pivot) in each row comes in a later column than the pivot in the previous row. All zero rows come last.

A factorization of the Fourier matrix Fn into e = log2 n matrices Si times a permutation. Each Si needs only nl2 multiplications, so Fnx and Fn-1c can be computed with ne/2 multiplications. Revolutionary.

Columns without pivots; these are combinations of earlier columns.

Entries HU = 1/(i + j -1) = Jd X i- 1 xj-1dx. Positive definite but extremely small Amin and large condition number: H is ill-conditioned.

Row n + 1 counts corners, edges, faces, ... of a cube in Rn.

The m by n edge-node incidence matrix has a row for each edge (node i to node j), with entries -1 and 1 in columns i and j .

Blocks aij B, eigenvalues Ap(A)Aq(B).

The subspace spanned by b, Ab, ... , Aj-Ib. Numerical methods approximate A -I b by x j with residual b - Ax j in this subspace. A good basis for K j requires only multiplication by A at each step.

A combination other than all Ci = 0 gives L Ci Vi = O.

Ps = pascal(n) = the symmetric matrix with binomial entries (i1~;2). Ps = PL Pu all contain Pascal's triangle with det = 1 (see Pascal in the index).

Columns that contain pivots after row reduction. These are not combinations of earlier columns. The pivot columns are a basis for the column space.

Iv·wl < IIvll IIwll.Then IvTAwl2 < (vT Av)(wT Aw) for pos def A.

The transpose is -K, since Kij = -Kji. Eigenvalues are pure imaginary, eigenvectors are orthogonal, eKt is an orthogonal matrix.

If x gives the movements of the nodes, K x gives the internal forces. K = ATe A where C has spring constants from Hooke's Law and Ax = stretching.

Constant down each diagonal = time-invariant (shift-invariant) filter.

For matrix norms II A + B II < II A II + II B II·