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PROBLEM 2. Consider the matrix A and three vectors v1, V2, and v3 given by [5 -1 A = 3 1 1 -1 -3 -3 Vi = V2 V3 = 1 1. Show that the three vectors v1, v2, and v3 form a basis for R. 2. Show that the three vectors are all eigenvectors for A and determine the eigenvalue corresponding to each eigenvector. Let a vector be given by 4 y = 3. Find the coordinates of y in the eigenvector basis {v1, V2; V3}. AFT PROBLEM 2. Solution The three vectors are a basis for R' if they are linearly independent and span R°. This can be checked by assembling the vectors in a matrix and row reducing. 1 1] [1 0 0] |1 0 1~ 0 1 0 [Vi v2 V3] 0 1 1 0 0 1 As the matrix is row equivalent to the identity matrix we can conclude that the vectors are indeed linearly independent and span R' and hence forms a basis for R. The second problem can be solved by direct computation 10-8 5 -1 -3 Avı 1 -3 = 4 1 -1 1 From the above it is seen that Avi = 4v1 and hence d1 = 4. Similar computations for V2 and v3 reveals that = 2 and Az 1. The coordinates of y in the eigenvector basis are the c;'s [v]. = 2 where y = ciV1+ ¢2V2 + €3V3 C3 This gives C1 3. y = [V1 V2 v3] 2 = [V1 V2 V3]y = = [v] C2 C2 2

PROBLEM 2. Consider the matrix A and three vectors v1, V2, and v3 given by [5 -1 A = 3 1 1 -1 -3 -3 Vi = V2 V3 = 1 1. Show that the three vectors v1, v2, and v3 form a basis for R. 2. Show that the three vectors are all eigenvectors for A and determine the eigenvalue corresponding to each eigenvector. Let a vector be given by 4 y = 3. Find the coordinates of y in the eigenvector basis {v1, V2; V3}. AFT PROBLEM 2. Solution The three vectors are a basis for R' if they are linearly independent and span R°. This can be checked by assembling the vectors in a matrix and row reducing. 1 1] [1 0 0] |1 0 1~ 0 1 0 [Vi v2 V3] 0 1 1 0 0 1 As the matrix is row equivalent to the identity matrix we can conclude that the vectors are indeed linearly independent and span R' and hence forms a basis for R. The second problem can be solved by direct computation 10-8 5 -1 -3 Avı 1 -3 = 4 1 -1 1 From the above it is seen that Avi = 4v1 and hence d1 = 4. Similar computations for V2 and v3 reveals that = 2 and Az 1. The coordinates of y in the eigenvector basis are the c;'s [v]. = 2 where y = ciV1+ ¢2V2 + €3V3 C3 This gives C1 3. y = [V1 V2 v3] 2 = [V1 V2 V3]y = = [v] C2 C2 2

Linear Algebra: A Modern Introduction
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter5: Orthogonality
Section5.3: The Gram-schmidt Process And The Qr Factorization
Problem 8AEXP
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Hello bartle

i have a question regarding eigenvector in the red circle in the solution thats what i need help for. i dont understand my teachers solution.  could you explain it step for step?

PROBLEM 2. Consider the matrix A and three vectors v1, V2, and v3 given by 5 -1 A = 3 1 1 -1 -3 -3 Vi = 1. Show that the three vectors v1, v2, and v3 form a basis for R. 2. Show that the three vectors are all eigenvectors for A and determine the eigenvalue corresponding to each eigenvector. Let a vector be given by 4 y = 3. Find the coordinates of y in the eigenvector basis {v1, V2, V3}. AFT PROBLEM 2. Solution The three vectors are a basis for R' if they are linearly independent and span R°. This can be checked by assembling the vectors in a matrix and row reducing. [1 0 0] 1 0 1 ~ 0 1 0 0 1 1 1 1] [Vi v2 V3] 0 0 1 As the matrix is row equivalent to the identity matrix we can conclude that the vectors are indeed linearly independent and span R' and hence forms a basis for R. The second problem can be solved by direct computation 5 -1 -3 Avi -3 = 4 1 -1 1 From the above it is seen that Avi V2 and v3 reveals that Ag = 2 and As = 1. 4v1 and hence A1 = 4. Similar computations for %3D The coordinates of y in the eigenvector basis are the c;'s [v]. = 2 where y = C1V1 + C2V2 + C3V3 C3 This gives C1 3. y = [V1 V2 V3] 2 [V1 V2 v3]y = = [v]. - C2 C2 C2
Transcribed Image Text:PROBLEM 2. Consider the matrix A and three vectors v1, V2, and v3 given by 5 -1 A = 3 1 1 -1 -3 -3 Vi = 1. Show that the three vectors v1, v2, and v3 form a basis for R. 2. Show that the three vectors are all eigenvectors for A and determine the eigenvalue corresponding to each eigenvector. Let a vector be given by 4 y = 3. Find the coordinates of y in the eigenvector basis {v1, V2, V3}. AFT PROBLEM 2. Solution The three vectors are a basis for R' if they are linearly independent and span R°. This can be checked by assembling the vectors in a matrix and row reducing. [1 0 0] 1 0 1 ~ 0 1 0 0 1 1 1 1] [Vi v2 V3] 0 0 1 As the matrix is row equivalent to the identity matrix we can conclude that the vectors are indeed linearly independent and span R' and hence forms a basis for R. The second problem can be solved by direct computation 5 -1 -3 Avi -3 = 4 1 -1 1 From the above it is seen that Avi V2 and v3 reveals that Ag = 2 and As = 1. 4v1 and hence A1 = 4. Similar computations for %3D The coordinates of y in the eigenvector basis are the c;'s [v]. = 2 where y = C1V1 + C2V2 + C3V3 C3 This gives C1 3. y = [V1 V2 V3] 2 [V1 V2 v3]y = = [v]. - C2 C2 C2
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