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 4.5.4.5.81: Proof Prove that if S = {v1, v2, . . . , vn} is a basisfor a vector...
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 4.5.4.5.82: Proof Prove Theorem 4.12.
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 4.5.4.5.85: Proof Let S be a linearly independent set of vectorsfrom a finite d...
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Solutions for Chapter 4.5: Basis and Dimension
Full solutions for Elementary Linear Algebra  8th Edition
ISBN: 9781305658004
Solutions for Chapter 4.5: Basis and Dimension
Get Full SolutionsThis expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Elementary Linear Algebra, edition: 8. Chapter 4.5: Basis and Dimension includes 172 full stepbystep solutions. Elementary Linear Algebra was written by and is associated to the ISBN: 9781305658004. Since 172 problems in chapter 4.5: Basis and Dimension have been answered, more than 44190 students have viewed full stepbystep solutions from this chapter.

Affine transformation
Tv = Av + Vo = linear transformation plus shift.

Column picture of Ax = b.
The vector b becomes a combination of the columns of A. The system is solvable only when b is in the column space C (A).

Commuting matrices AB = BA.
If diagonalizable, they share n eigenvectors.

Complete solution x = x p + Xn to Ax = b.
(Particular x p) + (x n in nullspace).

Cramer's Rule for Ax = b.
B j has b replacing column j of A; x j = det B j I det A

Diagonalizable matrix A.
Must have n independent eigenvectors (in the columns of S; automatic with n different eigenvalues). Then SI AS = A = eigenvalue matrix.

Dimension of vector space
dim(V) = number of vectors in any basis for V.

Ellipse (or ellipsoid) x T Ax = 1.
A must be positive definite; the axes of the ellipse are eigenvectors of A, with lengths 1/.JI. (For IIx II = 1 the vectors y = Ax lie on the ellipse IIA1 yll2 = Y T(AAT)1 Y = 1 displayed by eigshow; axis lengths ad

Fourier matrix F.
Entries Fjk = e21Cijk/n give orthogonal columns FT F = nI. Then y = Fe is the (inverse) Discrete Fourier Transform Y j = L cke21Cijk/n.

Free variable Xi.
Column i has no pivot in elimination. We can give the n  r free variables any values, then Ax = b determines the r pivot variables (if solvable!).

Full column rank r = n.
Independent columns, N(A) = {O}, no free variables.

Inverse matrix AI.
Square matrix with AI A = I and AAl = I. No inverse if det A = 0 and rank(A) < n and Ax = 0 for a nonzero vector x. The inverses of AB and AT are B1 AI and (AI)T. Cofactor formula (Al)ij = Cji! detA.

Jordan form 1 = M 1 AM.
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.

Left inverse A+.
If A has full column rank n, then A+ = (AT A)I AT has A+ A = In.

Linear combination cv + d w or L C jV j.
Vector addition and scalar multiplication.

Plane (or hyperplane) in Rn.
Vectors x with aT x = O. Plane is perpendicular to a =1= O.

Rank one matrix A = uvT f=. O.
Column and row spaces = lines cu and cv.

Row picture of Ax = b.
Each equation gives a plane in Rn; the planes intersect at x.

Singular Value Decomposition
(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.

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