 4.5.1E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.2E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.3E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.4E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.5E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.6E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.7E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.8E: For each subspace in Exercises 1–8, (a) find a basis for the subspa...
 4.5.9E: Find the dimension of the subspace of all vectors in whose first an...
 4.5.10E: Find the dimension of the subspace H of spanned by
 4.5.11E: In Exercises 11 and 12, find the dimension of the subspace spanned ...
 4.5.12E: In Exercises 11 and 12, find the dimension of the subspace spanned ...
 4.5.13E: Determine the dimensions of Nul A and Col A for the matrices shown ...
 4.5.14E: Determine the dimensions of Nul A and Col A for the matrices shown ...
 4.5.15E: Determine the dimensions of Nul A and Col A for the matrices shown ...
 4.5.16E: Determine the dimensions of Nul A and Col A for the matrices shown ...
 4.5.17E: Determine the dimensions of Nul A and Col A for the matrices shown ...
 4.5.18E: Determine the dimensions of Nul A and Col A for the matrices shown ...
 4.5.19E: In Exercises 19 and 20, V is a vector space. Mark each statement Tr...
 4.5.20E: In Exercises 19 and 20, V is a vector space. Mark each statement Tr...
 4.5.21E: The first four Hermite polynomials are 1, 2t, 2 + 4t2 and 12t + 8...
 4.5.22E: The first four Laguerre polynomials are 1, 1  t, 2  4t + t2, and ...
 4.5.23E: Let B be the basis of consisting of the Hermite polynomials in Exer...
 4.5.24E: Let be the basis of P2 consisting of the first three Laguerre polyn...
 4.5.25E: Let S be a subset of an ndimensional vector space V, and suppose S...
 4.5.26E: Let H be an ndimensional subspace of an ndimensional vector space...
 4.5.27E: Explain why the space of all polynomials is an infinitedimensional...
 4.5.28E: Show that the space of all continuous functions defined on the real...
 4.5.29E: In Exercises 29 and 30, V is a nonzero finitedimensional vector sp...
 4.5.30E: In Exercises 29 and 30, V is a nonzero finitedimensional vector sp...
 4.5.31E: Exercises 31 and 32 concern finitedimensional vector spaces V and ...
 4.5.32E: Exercises 31 and 32 concern finitedimensional vector spaces V and ...
 4.5.33E: [M] According to Theorem 11, a linearly independent set in can be e...
 4.5.34E: Assume the following trigonometric identities (see Exercise 37 in S...
Solutions for Chapter 4.5: Linear Algebra and Its Applications 4th Edition
Full solutions for Linear Algebra and Its Applications  4th Edition
ISBN: 9780321385178
Solutions for Chapter 4.5
Get Full SolutionsSince 34 problems in chapter 4.5 have been answered, more than 28124 students have viewed full stepbystep solutions from this chapter. Chapter 4.5 includes 34 full stepbystep solutions. This textbook survival guide was created for the textbook: Linear Algebra and Its Applications, edition: 4. Linear Algebra and Its Applications was written by and is associated to the ISBN: 9780321385178. This expansive textbook survival guide covers the following chapters and their solutions.

CayleyHamilton Theorem.
peA) = det(A  AI) has peA) = zero matrix.

Change of basis matrix M.
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.)

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).

Column space C (A) =
space of all combinations of the columns of 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).

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

Free columns of A.
Columns without pivots; these are combinations of earlier columns.

Hessenberg matrix H.
Triangular matrix with one extra nonzero adjacent diagonal.

Hilbert matrix hilb(n).
Entries HU = 1/(i + j 1) = Jd X i 1 xj1dx. Positive definite but extremely small Amin and large condition number: H is illconditioned.

lAII = l/lAI and IATI = IAI.
The big formula for det(A) has a sum of n! terms, the cofactor formula uses determinants of size n  1, volume of box = I det( A) I.

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

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

Norm
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.

Outer product uv T
= column times row = rank one matrix.

Partial pivoting.
In each column, choose the largest available pivot to control roundoff; all multipliers have leij I < 1. See condition number.

Polar decomposition A = Q H.
Orthogonal Q times positive (semi)definite H.

Reduced row echelon form R = rref(A).
Pivots = 1; zeros above and below pivots; the r nonzero rows of R give a basis for the row space of A.

Solvable system Ax = b.
The right side b is in the column space of A.

Spectrum of A = the set of eigenvalues {A I, ... , An}.
Spectral radius = max of IAi I.