Consider the matrix A= 1. Find the eigenvalues and eigenvectors of A. 2. Consider the polynomial p(r) = 5z? – 6z. Compute p(A). 3. For the two eigenvectors computed in the first part of the problem, show explicity that p(A)v = p(cv, if c is the corresponding eigenvalue to v.
Consider the matrix A= 1. Find the eigenvalues and eigenvectors of A. 2. Consider the polynomial p(r) = 5z? – 6z. Compute p(A). 3. For the two eigenvectors computed in the first part of the problem, show explicity that p(A)v = p(cv, 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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![Consider the matrix A =
3
1. Find the eigenvalues and eigenvectors of A.
2. Consider the polynomial p(z) = 5x? – 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, ifc 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, ..., n are eigenvectors to different eigenvalues c1, c2, ..., Cn. We want to show that these eigenvectors are linearly indepdendent To do so, write down a
linear combination of the form v = bịvi + bavat...+bnUn = 0 and show that all coefficient br are zero. To show, for example, that bị is zero, you can consider the polynomial
P1 (z) = (1 - 2)(x - c3)... (I - Cn) and evalutate pi (A)v.
%3!](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F44455328-4954-481c-8225-cdd51b7e94a1%2F9fd0525e-7a79-4bc1-8c30-59cb4ed859d9%2Fnstzv9w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Consider the matrix A =
3
1. Find the eigenvalues and eigenvectors of A.
2. Consider the polynomial p(z) = 5x? – 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, ifc 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, ..., n are eigenvectors to different eigenvalues c1, c2, ..., Cn. We want to show that these eigenvectors are linearly indepdendent To do so, write down a
linear combination of the form v = bịvi + bavat...+bnUn = 0 and show that all coefficient br are zero. To show, for example, that bị is zero, you can consider the polynomial
P1 (z) = (1 - 2)(x - c3)... (I - Cn) and evalutate pi (A)v.
%3!
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