2 Consider the matrix A = 1. Find the eigenvalues and eigenvectors of A. 2. Consider the polynomial p(r) = 5z2 – 6z. Compute p(A). 3. For the two eigenvectors computed in the first part of the problem, show explicitly that p(A)v = p(c)v, if c is the corresponding eigenvalue to v.
2 Consider the matrix A = 1. Find the eigenvalues and eigenvectors of A. 2. Consider the polynomial p(r) = 5z2 – 6z. Compute p(A). 3. For the two eigenvectors computed in the first part of the problem, show explicitly that p(A)v = p(c)v, if c is the corresponding eigenvalue to v.
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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Question
![2
Consider the matrix A =
1. Find the eigenvalues and eigenvectors of A.
2. Consider the polynomial p(z) = 522 – 6x. Compute p(A).
3. For the two eigenvectors computed in the first part of the problem, show explicitly that p(A jv = p(c)v, if c is the corresponding eigenvalue to v.
4. Show that p(A)v = p(c)v holds in general, if v is an eigenvector of A for the eigenvalue c.
5. Assume now that v1, v2, ..., Vn are eigenvectors to different eigenvalues cı, C2, ..., Cn. We want to show that these eigenvectors are linearly indepdendent. To do so, write down a
= 0 and show that all coefficient be are zero. To show, for example, that bị is zero, you can consider the polynomial
P1(A)v.
linear combination of the form v =
bivi + byvgt... +bnUn
P1 (2) = (r - c2)(z - c3)... (z - Cn) and evalutate](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44455328-4954-481c-8225-cdd51b7e94a1%2F7363a5ff-dc95-481b-bb24-70a28866ca2d%2F8hh7z85_processed.jpeg&w=3840&q=75)
Transcribed Image Text:2
Consider the matrix A =
1. Find the eigenvalues and eigenvectors of A.
2. Consider the polynomial p(z) = 522 – 6x. Compute p(A).
3. For the two eigenvectors computed in the first part of the problem, show explicitly that p(A jv = p(c)v, if c is the corresponding eigenvalue to v.
4. Show that p(A)v = p(c)v holds in general, if v is an eigenvector of A for the eigenvalue c.
5. Assume now that v1, v2, ..., Vn are eigenvectors to different eigenvalues cı, C2, ..., Cn. We want to show that these eigenvectors are linearly indepdendent. To do so, write down a
= 0 and show that all coefficient be are zero. To show, for example, that bị is zero, you can consider the polynomial
P1(A)v.
linear combination of the form v =
bivi + byvgt... +bnUn
P1 (2) = (r - c2)(z - c3)... (z - Cn) and evalutate
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