Math: Eigenvalues and Eigenvectors Worksheet

Question 1

For the matrix A = [[3, 0], [0, 5]], find the eigenvalues and one eigenvector for each eigenvalue.

Accepted Answer

The eigenvalues are 3 and 5. For eigenvalue 3, one eigenvector is [1, 0] because A[1, 0] = [3, 0]. For eigenvalue 5, one eigenvector is [0, 1] because A[0, 1] = [0, 5].

Question 2

For the matrix A = [[2, 1], [0, 2]], find the eigenvalue or eigenvalues.

Accepted Answer

The characteristic equation is det(A - λI) = det([[2 - λ, 1], [0, 2 - λ]]) = (2 - λ)^2 = 0. The only eigenvalue is λ = 2, with algebraic multiplicity 2.

Question 3

For the matrix A = [[4, 0], [0, -1]], determine whether v = [0, 3] is an eigenvector. If it is, give the eigenvalue.

Accepted Answer

Yes, v = [0, 3] is an eigenvector. Multiplying gives A[0, 3] = [0, -3], which equals -1[0, 3], so the eigenvalue is -1.

Question 4

Find the characteristic equation of A = [[1, 2], [3, 2]]. Then find the eigenvalues.

Accepted Answer

A - λI = [[1 - λ, 2], [3, 2 - λ]]. The determinant is (1 - λ)(2 - λ) - 6 = λ^2 - 3λ - 4. The characteristic equation is λ^2 - 3λ - 4 = 0, so the eigenvalues are λ = 4 and λ = -1.

Question 5

For A = [[1, 2], [3, 2]], find an eigenvector for the eigenvalue λ = 4.

Accepted Answer

For λ = 4, solve (A - 4I)v = 0. This gives [[-3, 2], [3, -2]][x, y] = [0, 0], so -3x + 2y = 0. One solution is x = 2 and y = 3, so one eigenvector is [2, 3].

Question 6

For A = [[1, 2], [3, 2]], find an eigenvector for the eigenvalue λ = -1.

Accepted Answer

For λ = -1, solve (A + I)v = 0. This gives [[2, 2], [3, 3]][x, y] = [0, 0], so x + y = 0. One solution is x = 1 and y = -1, so one eigenvector is [1, -1].

Question 7

The matrix A = [[0, 1], [1, 0]] swaps the coordinates of a vector, so A[x, y] = [y, x]. Show that [1, 1] and [1, -1] are eigenvectors, and give their eigenvalues.

Accepted Answer

A[1, 1] = [1, 1], so [1, 1] is an eigenvector with eigenvalue 1. A[1, -1] = [-1, 1], which equals -1[1, -1], so [1, -1] is an eigenvector with eigenvalue -1.

Question 8

A matrix transformation sends u = [2, 0] to [6, 0]. Is u an eigenvector of the transformation? Explain your answer and name the eigenvalue if there is one.

Accepted Answer

Yes, u is an eigenvector because its image [6, 0] is a scalar multiple of [2, 0]. Since [6, 0] = 3[2, 0], the eigenvalue is 3.

Question 9

For A = [[2, 0], [0, 2]], describe all nonzero eigenvectors and give the eigenvalue.

Accepted Answer

Every nonzero vector in the plane is an eigenvector of A. The matrix doubles every vector, so Av = 2v for any nonzero vector v. The eigenvalue is 2.

Question 10

A 2 by 2 matrix has characteristic equation λ^2 - 5λ + 6 = 0. Find its eigenvalues.

Accepted Answer

Factor the characteristic equation as λ^2 - 5λ + 6 = (λ - 2)(λ - 3) = 0. Therefore, the eigenvalues are λ = 2 and λ = 3.

Math: Eigenvalues and Eigenvectors

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