 5.4.1: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.2: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.3: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.4: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.5: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.6: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.7: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.8: In l:.xercises 18. a linear operator Ton n3 and a basis 8 for n3 a...
 5.4.9: n Exercises 9 20. a linear operator T on R " and its clwracterisri...
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 5.4.18: n Exercises 9 20. a linear oper ator T on R " and its clwracterisr...
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 5.4.20: n Exercises 9 20. a linear operator T on R " and its clwracterisri...
 5.4.21: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.22: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.23: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.24: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.25: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.26: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.27: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.28: In Exercises 21 28, a linear operator Ton n is given. Find, if pos...
 5.4.29: f a linear operator on n" is diagonalizable. then its standard matr...
 5.4.30: For every linear operator on n. there is a bttsis B for n such that...
 5.4.31: A linear operator on n is diagonalizablc if and only if its standar...
 5.4.32: If T is a diagonalizable linear operator on n. there is a unique ba
 5.4.33: lf T is a diagonalizable linear operator on n. there is a unique d
 5.4.34: et W be a 2dimcnsional subspace of n 3. The reflection of n
 5.4.35: Let W be a 2dimcnsional subspace of nJ The reflection of n 3 abou
 5.4.36: If T is a linear operator on n" and B is a basis for n such that [T...
 5.4.37: The characteristic polynomial of a linear operator T on 1?." is a poly
 5.4.38: If the characteristic polynomial of a linear operator T on 1?." fac...
 5.4.39: If the characteristic polynomial of a linear operator T on n" does ...
 5.4.40: If, for each eigenvalue A of a linear operator T on n, the dimensio...
 5.4.41: Let IV be a twodimensional subspace of n3 If Tw : n 3  n 3 is the...
 5.4.42: Let W be a twodimensional subspace of n 3. If Tw: n 3  n 3 is the...
 5.4.43: If T is a diagonalizable linear operator having 0 as an eigenvalu
 5.4.44: T is a linear operator on 1?.", then the sum of the multiplicitie
 5.4.45: If T is a diagonalizable linear operator on n", then the sum of the...
 5.4.46: If T is a linear operator on n having 11 distinct eigenvalues. then...
 5.4.47: If 8 1 8 2, . , Bk are bases for distinct eigenspaces of a linear op
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 5.4.49: n Exerci. 4958 inet1r operaTor tmd irs characrerisric polynomial a...
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 5.4.59: In Exercises 59 64, the equation of a plane W through the origin o...
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 5.4.63: In Exercises 59 64, the equation of a plane W through the origin o...
 5.4.64: In Exercises 59 64, the equation of a plane W through the origin o...
 5.4.65: In Exercises 65 and 66. the standard matrix of a reflection ofn3 ab...
 5.4.66: In Exercises 65 and 66. the standard matrix of a reflection ofn3 ab...
 5.4.67: Let IV be a 2dimensional subspace of n3 (a) Prove that there exist...
 5.4.68: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.69: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.70: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.71: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.72: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.73: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.74: In Exercises 6874, find an explicit fomwla for the orthogonal proj...
 5.4.75: Let [u. v. wf be a basis for n3 and let T be the linear operator on...
 5.4.76: Let (u. v. wf be a basis for n3 and let T be the linear operator on...
 5.4.77: Let T be a linear operator on n ru1d 8 be a basis for n" such that ...
 5.4.78: If T and U are diagonalizable linear operators on n", must T + U be...
 5.4.79: If T is a diagonalizable linear operator on n". must cT be a diagon...
 5.4.80: If T and U are diagonalizable linear operators o n 'R.", must TV be...
 5.4.81: Let T be a linear operator on n". and suppose that v1 v2 ..... Vt a...
 5.4.82: Let T and U be diagona lizable linear operators on n". Prove that i...
 5.4.83: Let T and U be linear operators on 'R.". If T 2 = U (where T 2 = 7T...
 5.4.84: et T be a li near operator on n and 6 1 62 ..... 8t be bases for a
 5.4.85: T is the linear operator on R 5 defi ned by [ x1] [I Lt1  9x2 + 1...
 5.4.86: T is the linear o perator on R 5 de fined by [ Xtl [ 2r,  4x2 9x...
Solutions for Chapter 5.4: EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION
Full solutions for Elementary Linear Algebra: A Matrix Approach  2nd Edition
ISBN: 9780131871410
Solutions for Chapter 5.4: EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION
Get Full SolutionsChapter 5.4: EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION includes 86 full stepbystep solutions. This textbook survival guide was created for the textbook: Elementary Linear Algebra: A Matrix Approach, edition: 2. This expansive textbook survival guide covers the following chapters and their solutions. Elementary Linear Algebra: A Matrix Approach was written by and is associated to the ISBN: 9780131871410. Since 86 problems in chapter 5.4: EIGENVALUES, EIGENVECTORS, AND DIAGONALIZATION have been answered, more than 25258 students have viewed full stepbystep solutions from this chapter.

Adjacency matrix of a graph.
Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected). Adjacency matrix of a graph. Square matrix with aij = 1 when there is an edge from node i to node j; otherwise aij = O. A = AT when edges go both ways (undirected).

Associative Law (AB)C = A(BC).
Parentheses can be removed to leave ABC.

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

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).

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

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

Iterative method.
A sequence of steps intended to approach the desired solution.

Linearly dependent VI, ... , Vn.
A combination other than all Ci = 0 gives L Ci Vi = O.

Minimal polynomial of A.
The lowest degree polynomial with meA) = zero matrix. This is peA) = det(A  AI) if no eigenvalues are repeated; always meA) divides peA).

Particular solution x p.
Any solution to Ax = b; often x p has free variables = o.

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

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.

Saddle point of I(x}, ... ,xn ).
A point where the first derivatives of I are zero and the second derivative matrix (a2 II aXi ax j = Hessian matrix) is indefinite.

Simplex method for linear programming.
The minimum cost vector x * is found by moving from comer to lower cost comer along the edges of the feasible set (where the constraints Ax = b and x > 0 are satisfied). Minimum cost at a comer!

Special solutions to As = O.
One free variable is Si = 1, other free variables = o.

Symmetric matrix A.
The transpose is AT = A, and aU = a ji. AI is also symmetric.

Trace of A
= sum of diagonal entries = sum of eigenvalues of A. Tr AB = Tr BA.

Transpose matrix AT.
Entries AL = Ajj. AT is n by In, AT A is square, symmetric, positive semidefinite. The transposes of AB and AI are BT AT and (AT)I.

Tridiagonal matrix T: tij = 0 if Ii  j I > 1.
T 1 has rank 1 above and below diagonal.