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Directions a matrix only scales

Mathematics · Linear Algebra · ages 19-20

Name ______________________ Date ____________

For A = [[0, 0], [0, 4]], v = (1, 0) is an eigenvector. Which calculation gives its eigenvalue?
- A v = (1, 0), so lambda = 1.
- A v = (0, 0), so lambda = 0.
- A v = (4, 0), so lambda = 4.
- A v = (0, 4), so lambda = 4.
The matrix A = [[2, 0], [0, 3]] is diagonal. What are its eigenvalues?
- 2 and 3
- 5 and 1
- 0 and 5
Take v = (1, 1) and A = [[1, 2], [0, 3]]. Is v an eigenvector of A?
Circle one: True False
Which equation do you solve to find the eigenvalues of a square matrix A?
- A v = 0
- det(A) = lambda
- det(A minus lambda I) = 0
The matrix [[0, minus 1], [1, 0]] rotates the plane by 90 degrees. Does it have any real eigenvalues?
- Yes: 1 and minus 1
- No, no real roots
- Yes: 0
A = [[-1, 0], [0, -1]] is a half turn about the origin. Which statement describes its real eigenvectors?
- Only nonzero vectors along the x-axis are eigenvectors; their eigenvalue is -1.
- Only nonzero vectors along the y-axis are eigenvectors; their eigenvalue is -1.
- No nonzero real vector is an eigenvector; the turn preserves no real line.
- Every nonzero real vector is an eigenvector; its eigenvalue is -1.
For A = [[4, 2], [0, 1]] and lambda = 1, A minus I = [[3, 2], [0, 0]]. An eigenvector is (2, k). What is k?
Answer: ______________
A = [[4, 1], [2, 3]] has characteristic polynomial lambda squared minus 7 lambda + 10. What is the larger eigenvalue?
Answer: ______________
Pat reports the zero vector as an eigenvector. What is wrong?
- Zero never qualifies
- Zero qualifies whenever A v = 0
- Only diagonal matrices reject zero
A = [[5, 4], [1, 2]]. Form the characteristic polynomial. What is the smaller eigenvalue?
Answer: ______________

Answer key

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Directions a matrix only scales W1-mt_TVqqaw11qa-s1

A v = (0, 0), so lambda = 0. · A(1, 0) = (0, 0) = 0(1, 0), so the eigenvalue is 0. The eigenvector itself is nonzero.
2 and 3 · Diagonal entries are the eigenvalues: 2 and 3.
True · Yes: A v = (3, 3) = 3v, so v is an eigenvector with eigenvalue 3.
det(A minus lambda I) = 0 · Nontrivial solutions of (A minus lambda I)v = 0 need a zero determinant.
No, no real roots · Every nonzero vector turns, so no real direction is preserved; the polynomial confirms it.
Every nonzero real vector is an eigenvector; its eigenvalue is -1. · For every vector v = (x, y), A v = (-x, -y) = -v. Every nonzero vector is therefore an eigenvector with eigenvalue -1.
-3 · Row 1 gives 3x + 2y = 0, so with x = 2, y = -3. Check: A(2, -3) = (8 - 6, -3) = (2, -3), which is 1 times v.
5 · Factor: (lambda minus 5)(lambda minus 2) = 0, so the larger root is 5.
Zero never qualifies · An eigenvector must be nonzero. The zero vector is excluded by the definition.
1 · det(A minus lambda I) = (5 minus lambda)(2 minus lambda) minus 4 = lambda squared minus 7 lambda + 6 = (lambda minus 6)(lambda minus 1), so the smaller eigenvalue is 1.