Explanation Given: The linear transformation T : V → V is defined by, T [ a b c d ] = [ − 3 a + 5 d 3 b − 5 c − 2 c 2 d ] The set is C = { A 1 , A 2 , A 3 , A 4 } , Here, A 1 = [ 1 0 0 1 ] , A 2 = [ 0 1 1 0 ] A 3 = [ 0 1 0 0 ] and A 4 = [ 1 0 0 0 ] Approach: If all the four given vectors A 1 , A 2 , A 3 and A 4 are linearly independent and if dim ( V ) = 4 . Then C = { A 1 , A 2 , A 3 , A 4 } is a basis of V . If T : X → X is a linear transformation then a scalar k is called the eigenvalue if there exist a non zero vector x such that T ( x ) = k x and the vector x is called the eigenvector. Let T : U → V be a linear transformation with dim ( U ) = n and dim ( V ) = m . Let B = { u 1 , u 2 , ... , u n } be a basis for U , and let C = { v 1 , v 2 , ... , v n } be a basis for V . The matrix representation for T with respect to the bases B and C is the ( m × n ) matrix Q defined by Q = [ Q 1 , Q 2 , ... , Q n ] where Q j = [ T ( u j ) ] C . Calculation: For C = { A 1 , A 2 , A 3 , A 4 } to be linearly independent, a A 1 + b A 2 + c A 3 + d A 4 = 0 a [ 1 0 0 1 ] + b [ 0 1 1 0 ] + c [ 0 1 0 0 ] + d [ 1 0 0 0 ] = [ 0 0 0 0 ] Compare the components, a + d = 0 b + c = 0 b = 0 a = 0 Thus, a = 0 , b = 0 , c = 0 and d = 0 . Hence, C = { A 1 , A 2 , A 3 , A 4 } is linearly independent and dim ( V ) = 4 . Therefore, C = { A 1 , A 2 , A 3 , A 4 } is a basis of V . To find T ( A 1 ) , T ( A 2 ) , T ( A 3 ) and T ( A 4 ) T ( A 1 ) = T ( [ 1 0 0 1 ] ) = [ 2 0 0 2 ] = 2 [ 1 0 0 1 ] = 2 ⋅ A 1 = 2 ⋅ A 1 + 0 ⋅ A 2 + 0 ⋅ A 3 + 0 ⋅ A 4 Thus, [ T ( A 1 ) ] C = [ 2 0 0 0 ]

Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

5th Edition

ISBN: 9780134689531

Author: Lee Johnson, Dean Riess, Jimmy Arnold

Publisher: PEARSON

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Chapter 5.10, Problem 3E

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That C={A1,A2,A3,A4} is a basis of V consisting of eigenvectors for T. Also, calculate the matrix of T with respect to C.

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Chapter 5.10, Problem 2E

Chapter 5.10, Problem 4E

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Introduction to Linear Algebra (Classic Version) (5th Edition) (Pearson Modern Classics for Advanced Mathematics Series)

Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - For u,v and w given in given Exercises 13,...Ch. 5.2 - Prob. 4E Ch. 5.2 - Prob. 5E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - Prob. 7E Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...Ch. 5.2 - In Exercises 6-11, the given set is a subset of a...

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That C = { A 1 , A 2 , A 3 , A 4 } is a basis of V consisting of eigenvectors for T . Also, calculate the matrix of T with respect to C .

Solution Summary: The author explains that the linear transformation T:Vto V is defined by, C=leftA_1,

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To show:

That C={A1,A2,A3,A4} is a basis of V consisting of eigenvectors for T. Also, calculate the matrix of T with respect to C.

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That C = { A 1 , A 2 , A 3 , A 4 } is a basis of V consisting of eigenvectors for T . Also, calculate the matrix of T with respect to C .

Solution Summary: The author explains that the linear transformation T:Vto V is defined by, C=leftA_1,

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To show: That C={A1,A2,A3,A4} is a basis of V consisting of eigenvectors for T. Also, calculate the matrix of T with respect to C.

Want to see the full answer?

Chapter 5 Solutions

To show:

That C={A1,A2,A3,A4} is a basis of V consisting of eigenvectors for T. Also, calculate the matrix of T with respect to C.