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Consider A = Го о 1 00 11 10000 0 01 0 1 0 0 0 0-1 0 1 -1 0 107000 0 TOMOTO The characteristic polynomial of A is r³(r − 1) (you don't have to verify that). (a) Recall that a nonzero v in Null(A - XI) is called an eigenvector of A. A generalized eigenvector is a nonzero vector in Null(A - XI)" for some n. Calculate A², A³. Find the generalized eigenvectors of A. (Hint: the dimension of a generalized eigenspace is equal to the algebraic multiplicity of the corresponding eigenvalue.) (b) Find a nonzero vector v in Null(A)n Col(A²). Find a vector u in Null(A) nCol(A) linearly independent from v. (c) Find a vector w such that v = A²w and a vector z such that u = Az. Conclude that w, Aw, A²w, z, Az form a basis for the generalized eigenspace corresponding to λ = 0. (Remark: We can always find a similar basis in general. How many linearly independent eigenvectors we should look for in the column space of (A - XI)" for each n depends on the dimensions of Null(A - XI)" for different ns. You can read about Jordan Canonical forms if interested how to do that).

Consider A = Го о 1 00 11 10000 0 01 0 1 0 0 0 0-1 0 1 -1 0 107000 0 TOMOTO The characteristic polynomial of A is r³(r − 1) (you don't have to verify that). (a) Recall that a nonzero v in Null(A - XI) is called an eigenvector of A. A generalized eigenvector is a nonzero vector in Null(A - XI)" for some n. Calculate A², A³. Find the generalized eigenvectors of A. (Hint: the dimension of a generalized eigenspace is equal to the algebraic multiplicity of the corresponding eigenvalue.) (b) Find a nonzero vector v in Null(A)n Col(A²). Find a vector u in Null(A) nCol(A) linearly independent from v. (c) Find a vector w such that v = A²w and a vector z such that u = Az. Conclude that w, Aw, A²w, z, Az form a basis for the generalized eigenspace corresponding to λ = 0. (Remark: We can always find a similar basis in general. How many linearly independent eigenvectors we should look for in the column space of (A - XI)" for each n depends on the dimensions of Null(A - XI)" for different ns. You can read about Jordan Canonical forms if interested how to do that).

Elementary Linear Algebra (MindTap Course List)
Elementary Linear Algebra (MindTap Course List)
8th Edition
ISBN:9781305658004
Author:Ron Larson
Publisher:Ron Larson
Chapter7: Eigenvalues And Eigenvectors
Section7.1: Eigenvalues And Eigenvectors
Problem 8E
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2. Consider A - Го о 00 1 1 00 01 01 1 0 1 0 0 0 0 0 1 -1 1 -1 0 0 0 1 0 0 0 -1 00 0 The characteristic polynomial of A is r5 (r - 1) (you don't have to verify that). (a) Recall that a nonzero v in Null(A - XI) is called an eigenvector of A. A generalized eigenvector is a nonzero vector in Null(A – XI)" for some n. Calculate A², A³. Find the generalized eigenvectors of A. (Hint: the dimension of a generalized eigenspace is equal to the algebraic multiplicity of the corresponding eigenvalue.) (b) Find a nonzero vector v in Null(A)n Col(A²). Find a vector u in Null(A)n Col(A) linearly independent from v. (c) Find a vector w such that v A²w and a vector z such that u = Az. Conclude that w, Aw, A²w, z, Az form a basis for the generalized eigenspace corresponding to λ = 0. (Remark: We can always find a similar basis in general. How many linearly independent eigenvectors we should look for in the column space of (A - XI)" for each n depends on the dimensions of Null(A - XI)" for different ns. You can read about Jordan Canonical forms if interested how to do that).
Transcribed Image Text:2. Consider A - Го о 00 1 1 00 01 01 1 0 1 0 0 0 0 0 1 -1 1 -1 0 0 0 1 0 0 0 -1 00 0 The characteristic polynomial of A is r5 (r - 1) (you don't have to verify that). (a) Recall that a nonzero v in Null(A - XI) is called an eigenvector of A. A generalized eigenvector is a nonzero vector in Null(A – XI)" for some n. Calculate A², A³. Find the generalized eigenvectors of A. (Hint: the dimension of a generalized eigenspace is equal to the algebraic multiplicity of the corresponding eigenvalue.) (b) Find a nonzero vector v in Null(A)n Col(A²). Find a vector u in Null(A)n Col(A) linearly independent from v. (c) Find a vector w such that v A²w and a vector z such that u = Az. Conclude that w, Aw, A²w, z, Az form a basis for the generalized eigenspace corresponding to λ = 0. (Remark: We can always find a similar basis in general. How many linearly independent eigenvectors we should look for in the column space of (A - XI)" for each n depends on the dimensions of Null(A - XI)" for different ns. You can read about Jordan Canonical forms if interested how to do that).
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