Math: Eigenvalues and Eigenvectors
Finding special directions and scale factors for matrices
Math: Eigenvalues and Eigenvectors
Finding special directions and scale factors for matrices
Math - Grade 9-12
- 1
For the matrix A = [[3, 0], [0, 5]], find the eigenvalues and one eigenvector for each eigenvalue.
A diagonal matrix stretches the x and y directions by the diagonal entries.
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]. - 2
For the matrix A = [[2, 1], [0, 2]], find the eigenvalue or eigenvalues.
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. - 3
For the matrix A = [[4, 0], [0, -1]], determine whether v = [0, 3] is an eigenvector. If it is, give the eigenvalue.
A vector is an eigenvector if Av is a scalar multiple of v.
Yes, v = [0, 3] is an eigenvector. Multiplying gives A[0, 3] = [0, -3], which equals -1[0, 3], so the eigenvalue is -1. - 4
Find the characteristic equation of A = [[1, 2], [3, 2]]. Then find the eigenvalues.
For a 2 by 2 matrix [[a, b], [c, d]], the determinant is ad - bc.
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. - 5
For A = [[1, 2], [3, 2]], find an eigenvector for the eigenvalue λ = 4.
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]. - 6
For A = [[1, 2], [3, 2]], find an eigenvector for the eigenvalue λ = -1.
Substitute λ = -1 into A - λI. Since subtracting -1 is adding 1, use A + I.
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]. - 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.
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. - 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.
Compare the output vector to the input vector. They must point in the same or opposite direction for an eigenvector.
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. - 9
For A = [[2, 0], [0, 2]], describe all nonzero eigenvectors and give the eigenvalue.
This matrix scales the whole plane by the same factor in every direction.
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. - 10
A 2 by 2 matrix has characteristic equation λ^2 - 5λ + 6 = 0. Find its eigenvalues.
Factor the characteristic equation as λ^2 - 5λ + 6 = (λ - 2)(λ - 3) = 0. Therefore, the eigenvalues are λ = 2 and λ = 3.