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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).

Advanced Engineering Mathematics
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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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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