Introduction to Linear Algebra 4th Edition - Solutions by Chapter

Ax = b is solvable when b is in the column space of A; then [A b] has the same rank as A. Elimination on [A b] keeps equations correct.

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!

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

Remove row i and column j; multiply the determinant by (-I)i + j •

If diagonalizable, they share n eigenvectors.

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.

The nullspace N (A) and row space C (AT) are orthogonal complements in Rn(perpendicular from Ax = 0 with dimensions rand n - r). Applied to AT, the column space C(A) is the orthogonal complement of N(AT) in Rm.

Constant along each antidiagonal; hij depends on i + j.

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.

All mij > 0 and each column sum is 1. Largest eigenvalue A = 1. If mij > 0, the columns of Mk approach the steady state eigenvector M s = s > O.

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.

= column times row = rank one matrix.

Orthogonal Q times positive (semi)definite H.

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.

(SVD) A = U:E VT = (orthogonal) ( diag)( orthogonal) First r columns of U and V are orthonormal bases of C (A) and C (AT), AVi = O'iUi with singular value O'i > O. Last columns are orthonormal bases of nullspaces.

Combinations of VI, ... ,Vm fill the space. The columns of A span C (A)!

v + w = (VI + WI, ... , Vn + Wn ) = diagonal of parallelogram.

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.