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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad6d3522-58fe-4a42-a237-8062bf310251%2Ff0e67889-7242-4406-aba3-428b8978ade9%2Fpadx6tp_processed.png&w=3840&q=75)
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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