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Working over Q, let a be the linear map corresponding to the matrix A = -1 2 4 with respect to the standard basis of V = Q³. Find the subspace 2 1 0 E(3, a). Suppose we are told that if A3 is a rational number then X is not an eigenvalue. Is a diagonalisable?

Working over Q, let a be the linear map corresponding to the matrix A = -1 2 4 with respect to the standard basis of V = Q³. Find the subspace 2 1 0 E(3, a). Suppose we are told that if A3 is a rational number then X is not an eigenvalue. Is a diagonalisable?

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 0 -1 4 -1 2 2 1 0 E(3, a). Suppose we are told that if X 3 is a rational number then X is not an eigenvalue. Is a diagonalisable? Working over Q, let a be the linear map corresponding to the matrix A = with respect to the standard basis of V=Q³. Find the subspace Select one: O Solving the linear equations for the entries of a column vector to form an eigenvector with eigenvalue 3 gives three linearly independent eigenvectors. Hence E(3, a) is 3-dimensional, which is also the dimension of V. Hence E(3, a) = V and moreover we see that a is diagonalisable O None of the others apply O The question does not say that 3 is an eigenvalue and indeed it is not, so E(3, a) = {0}. If there are no other possible eigenvalues over Q then we cannot find a basis of eigenvectors and a is not diagonalisable O O Solving the linear equations for the entries of a column vector to form an eigenvector with eigenvalue 2 gives v₁ = 8--8 hence E(3, a) = (v1, v2), which is 2-dimensional. As we are told that this is the only eigenvalue, the map is not diagonalisable as Vis 3-dimensional. Solving the linear equations for the entries of a column vector to form an eigenvector with eigenvalue 3 gives, up to scaling, only ₁ And 1 E(3, a) = (v₁), which is 1-dimensional. As we are told that this is the only eigenvalue over Q, the map is not diagonalisable as Vis 3-dimensional. as eigenvectors and 5 and hence
Transcribed Image Text:2 0 -1 4 -1 2 2 1 0 E(3, a). Suppose we are told that if X 3 is a rational number then X is not an eigenvalue. Is a diagonalisable? Working over Q, let a be the linear map corresponding to the matrix A = with respect to the standard basis of V=Q³. Find the subspace Select one: O Solving the linear equations for the entries of a column vector to form an eigenvector with eigenvalue 3 gives three linearly independent eigenvectors. Hence E(3, a) is 3-dimensional, which is also the dimension of V. Hence E(3, a) = V and moreover we see that a is diagonalisable O None of the others apply O The question does not say that 3 is an eigenvalue and indeed it is not, so E(3, a) = {0}. If there are no other possible eigenvalues over Q then we cannot find a basis of eigenvectors and a is not diagonalisable O O Solving the linear equations for the entries of a column vector to form an eigenvector with eigenvalue 2 gives v₁ = 8--8 hence E(3, a) = (v1, v2), which is 2-dimensional. As we are told that this is the only eigenvalue, the map is not diagonalisable as Vis 3-dimensional. Solving the linear equations for the entries of a column vector to form an eigenvector with eigenvalue 3 gives, up to scaling, only ₁ And 1 E(3, a) = (v₁), which is 1-dimensional. As we are told that this is the only eigenvalue over Q, the map is not diagonalisable as Vis 3-dimensional. as eigenvectors and 5 and hence
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