California Algebra 2: Concepts, Skills, and Problem Solving 1st Edition - Solutions by Chapter

The old basis vectors v j are combinations L mij Wi of the new basis vectors. The coordinates of CI VI + ... + cnvn = dl wI + ... + dn Wn are related by d = M c. (For n = 2 set VI = mll WI +m21 W2, V2 = m12WI +m22w2.)

Constant diagonals wrap around as in cyclic shift S. Every circulant is Col + CIS + ... + Cn_lSn - l . Cx = convolution c * x. Eigenvectors in F.

A sequence of steps (end of Chapter 9) to solve positive definite Ax = b by minimizing !x T Ax - x Tb over growing Krylov subspaces.

S. Permutation with S21 = 1, S32 = 1, ... , finally SIn = 1. Its eigenvalues are the nth roots e2lrik/n of 1; eigenvectors are columns of the Fourier matrix F.

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.

If A has s independent eigenvectors, its "generalized" eigenvector matrix M gives 1 = diag(lt, ... , 1s). The block his Akh +Nk where Nk has 1 's on diagonall. Each block has one eigenvalue Ak and one eigenvector.

The vector x that minimizes the error lie 112 solves AT Ax = ATb. Then e = b - Ax is orthogonal to all columns of A.

Nullspace of AT = "left nullspace" of A because y T A = OT.

The pivot row j is multiplied by eij and subtracted from row i to eliminate the i, j entry: eij = (entry to eliminate) / (jth pivot).

A directed graph that has constants Cl, ... , Cm associated with the edges.

IIA II. The ".e 2 norm" of A is the maximum ratio II Ax II/l1x II = O"max· Then II Ax II < IIAllllxll and IIABII < IIAIIIIBII and IIA + BII < IIAII + IIBII. Frobenius norm IIAII} = L La~. The.e 1 and.e oo norms are largest column and row sums of laij I.

If N NT = NT N, then N has orthonormal (complex) eigenvectors.

Projection p = P b is the closest point to b in S, error e = b - Pb is perpendicularto S. p 2 = P = pT, eigenvalues are 1 or 0, eigenvectors are in S or S...L. If columns of A = basis for S then P = A (AT A) -1 AT.

If A has full row rank m, then A+ = AT(AAT)-l has AA+ = 1m.

Real symmetric A has real A'S and orthonormal q's.

The transpose is AT = A, and aU = a ji. A-I is also symmetric.

T- 1 has rank 1 above and below diagonal.

Orthonormal columns (complex analog of Q).

Sequence of n real numbers v = (VI, ... , Vn) = point in Rn.

The rows (or the columns) of A generate a box with volume I det(A) I.