**Unformatted text preview: **2017-6-6 UW Common Math 308 Section 6.1 YIRAN CHEN
Math 308, section G, Spring 2017
Instructor: Amos Turchet WebAssign UW Common Math 308 Section 6.1 (Homework)
Current Score : 30 / 30 Due : Thursday, May 18 2017 11:00 PM PDT The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your
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Request Extension 1. 3/3 points | Previous AnswersHoltLinAlg1 6.1.002. Determine which of x1, x2, and x3 is an eigenvector for the matrix A. For those that are, determine
the associated eigenvalue. (For each vector, enter the associated eigenvalue, if it exists. If an
eigenvalue does not exist, enter DNE.)
A = −1 2
0 3 , x1 = 0
3 , x2 = 1
4 , x3 = 2
4 $$DNE
x1 $$DNE x2 $$3 x3 Solution or Explanation
Ax1 = Ax2 = Ax3 = −1 2 0 0 3 3 −1 2 1 0 3 4 −1 2 2 0 3 4 = = = 6
9
7
12
6
12 ≠ λx1 for any λ, so x1 is not an eigenvector. ≠ λx2 for any λ, so x2 is not an eigenvector. = 3 2
4 = 3x3, so x3 is an eigenvector with associated eigenvalue λ = 3. 1/11 2017-6-6 UW Common Math 308 Section 6.1 2. 3/3 points | Previous AnswersHoltLinAlg1 6.1.004. Determine which of x1, x2, and x3 is an eigenvector for the matrix A. For those that are,
determine the associated eigenvalue. (For each vector, enter the associated eigenvalue, if it exists. If
an eigenvalue does not exist, enter DNE.)
3 −1 0
A = 1 1 1
2 −1 3 0 , x1 = 1 , x2 = 1 , x3 = −1 1 2 1 0 −1 $$2
x1 $$2 x2 $$DNE x3 Solution or Explanation
3 −1 0
Ax1 = 1 2 1 −1 3 0 1 = 2 = 2 1 = 2x1, so x1 is an eigenvector with associated −1 1 2 1 2 1 1 2 1 eigenvalue λ = 2. 3 −1 0
Ax2 = −1 3 0 1 = 2 = 2 1 = 2x2, so x2 is an eigenvector with associated −1 1 2 0 0 0 eigenvalue λ = 2. 3 −1 0
Ax3 = −1 3 0 −1 1 2 1 1 2 = 5 ≠ λx3 for any λ, so x3 is not an eigenvector. −1 −1 2/11 2017-6-6 UW Common Math 308 Section 6.1 3. 1/1 points | Previous AnswersHoltLinAlg1 6.1.014. Find a basis for the eigenspace of A associated with the given eigenvalue λ.
A = −11 12
, λ = −3
−8 9 3/2
1 [3/2;1] Solution or Explanation
We rowreduce to obtain the null space of A − (−3)I2 = 3
2 x = s . A basis for the λ = −3 eigenspace is 1 3
2 −8 12
−8 12 ~ . Solving, we obtain −8 12
0 0
. 1 4. 1/1 points | Previous AnswersHoltLinAlg1 6.1.015. Find a basis for the eigenspace of A associated with the given eigenvalue λ.
8 −3 5
A = 8 1 1 , λ = 4 8 −3 5
1
3
1 [1;3;1] Solution or Explanation
We rowreduce to obtain the null space of A − 4I3 = 4 −3 5 4 −3 8 −3 1 ~ 0 3 −9 . Solving, we 8 −3 1 0 0 1
obtain x = s 3 . A basis for the λ = 4 eigenspace is 1 5
0 1
3 . 1 3/11 2017-6-6 UW Common Math 308 Section 6.1 5. 4/4 points | Previous AnswersHoltLinAlg1 6.1.024. Consider the matrix A.
A = −5 20
1 −4 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) $$λ2+9λ Find the real eigenvalues for the matrix A. (Enter your answers as a commaseparated list.) λ =
$$0, −9 Find a basis for each eigenspace for the matrix A. -5
1 (smaller eigenvalue) [5;1] 4
1 (larger eigenvalue) [4;1] Solution or Explanation
Characteristic polynomial: −5 20
1 0
det(A − λI2) = det − λ
1 −4
0 1
Eigenvalues: λ2 + 9λ = λ(λ + 9) = 0 = det 20
−λ − 5
1
−λ − 4 = λ2 + 9λ. λ = 0 or λ = −9. Eigenspace of λ = −9:
4 20
4 20 ~ ,
1 5
0 0
−5
so a basis for this eigenspace is . 1
A − (−9)I2 = Eigenspace of λ = 0:
−5 20
−5 20 ~ ,
1 −4
0 0
4
so a basis for this eigenspace is .
1
A − 0I2 = 4/11 2017-6-6 UW Common Math 308 Section 6.1 6. 5/5 points | Previous AnswersHoltLinAlg1 6.1.025. Consider the matrix A.
A = 4 0 0 1 3 0 −6 7 −1
Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) $$(4−λ)(3−λ)(−1−λ) Find the real eigenvalues for the matrix A. (Enter your answers as a commaseparated list.) λ =
$$4, 3, −1 Find a basis for each eigenspace for the matrix A. 0
0 (smallest eigenvalue) 1 [0;0;1] [0;4/7;1] 0
4/7
1 5
5 (largest eigenvalue) 1 [5;5;1] Solution or Explanation
Characteristic polynomial: det(A − λI3) = det 4 0 0 1 0 0 1 3 0 − λ 0 1 0 −6 7 −1 0 0 1 = det 4 − λ 0 0 1 3 − λ 0 −6 7 −1 − λ = −(λ − 3)(λ − 4)(λ + 1).
Eigenvalues: −(λ − 3)(λ − 4)(λ + 1) = 0 λ = 3, λ = 4, and λ = −1. Eigenspace of λ = 3: 5/11 2017-6-6 UW Common Math 308 Section 6.1 A − 3I3 = 1 0 0 1 0 1 0 0 ~ 0 7 −4 , −6 7 −4 0 0 0 0 0
so a basis for this eigenspace is 4
7 . 1
Eigenspace of λ = 4:
0
A − 4I3 = 0 1 −1
−6 0 1 −1 0 ~ 0 1 −5 , 0 0 7 −5 0
0 5
so a basis for this eigenspace is 5 . 1
Eigenspace of λ = −1:
A − (−1)I3 = 5 0 0 5 0 0 1 4 0 ~ 0 4 0 , −6 7 0 0 0 0
0 so a basis for this eigenspace is 0 . 1 6/11 2017-6-6 UW Common Math 308 Section 6.1 7. 3/3 points | Previous AnswersHoltLinAlg1 6.1.026. Consider the matrix A.
0 0 1
A = 1 0 0
0 1 0 Find the characteristic polynomial for the matrix A. (Write your answer in terms of λ.) $$−λ3+1 Find the real eigenvalues for the matrix A. (Enter your answers as a commaseparated list.) λ =
$$1 Find a basis for each eigenspace for the matrix A. 1
1
1 [1;1;1] Solution or Explanation
Characteristic polynomial: det(A − λI3) = det 0 0 1 1 0 0 1 0 0 − λ 0 1 0 0 1 0 0 0 1 = det Eigenvalues: −λ3 + 1 = −(λ − 1)(λ2 + λ + 1) = 0 −λ 0 1 1 −λ 0 0 1 −λ = −λ3 + 1. λ = 1. Eigenspace of λ = 1:
−1
A − (1)I3 = 0 1 −1
0 1 −1 0 ~ 0 1 0 −1 1 , 1 −1 0 0 0 1
so a basis for this eigenspace is 1 . 1 7/11 2017-6-6 UW Common Math 308 Section 6.1 8. 1/1 points | Previous AnswersHoltLinAlg1 6.1.037. Determine if the statement is true or false, and justify your answer.
An eigenvalue λ must be nonzero, but an eigenvector u can be equal to the zero vector.
True. This is part of the definition of multiplicity.
True. This is part of the definition of eigenvalues and eigenvectors. False. An eigenvalue must be nonzero, and an eigenvector must be a nonzero vector.
False. An eigenvalue may be 0, and an eigenvector may be equal to the zero vector. False. An eigenvalue may be 0, and an eigenvector must be a nonzero vector. Solution or Explanation
1 0 which has eigenvalues λ = 0 and 0 0
λ = 1. Moreover, by this definition, an eigenvector must be a nonzero vector. False. An eigenvalue may be 0, as with the matrix A = 9. 1/1 points | Previous AnswersHoltLinAlg1 6.1.039. Determine if the statement is true or false, and justify your answer.
If u is a nonzero eigenvector of A, then u and Au point in the same direction.
True. Since u is a nonzero eigenvector of A, there exists λ < 0 such that Au = λu.
True. Since u is a nonzero eigenvector of A, there exists λ > 0 such that Au = λu. False. Au and u are perpendicular. False. If λ < 0, then Au and u point in opposite directions.
False. If λ > 0, then Au and u point in opposite directions. Solution or Explanation
False, if λ < 0, then Au and u point in opposite directions. For instance, if A = u = −1 0
0 0 and 1
, then Au = −u.
0 8/11 2017-6-6 UW Common Math 308 Section 6.1 10.1/1 points | Previous AnswersHoltLinAlg1 6.1.043. Determine if the statement is true or false, and justify your answer.
If 0 is an eigenvalue of A, then nullity(A) > 0. True. Since 0 is an eigenvalue, there exists a nonzero vector x such that Ax = 0, and thus
nullity(A) > 0.
True. Since 0 is an eigenvalue, A is onetoone, and thus nullity(A) > 0. False. Since 0 is an eigenvalue, there exists a nonzero vector x such that Ax = 0, and thus
nullity(A) = 0.
01
False. Consider .
10
10
False. Consider .
01 Solution or Explanation
True. Since 0 is an eigenvalue, by the Big Theorem there exists a nonzero vector x such that Ax = 0, and thus nullity(A) > 0. 11.1/1 points | Previous AnswersHoltLinAlg1 6.1.044. Determine if the statement is true or false, and justify your answer.
Row operations do not change the eigenvalues of a matrix.
True. Since row operations do not change the determinant, the eigenvalues are unchanged.
True. Since row operations do not change the row space, the eigenvalues are unchanged. 10
01
False. Consider , which has eigenvalues 0 and 1, and , which has eigenvalue 0.
00
00
10
00 False. Consider , which has eigenvalues 0 and 1, and , which has eigenvalue 0.
00
10
01
02
False. Consider , which has eigenvalues 0 and 1, and , which has eigenvalues 0 and 2.
00
00 Solution or Explanation
False. The eigenvalues of an eigenvalue of 1 0
0 0 are 0 and 1. But upon interchanging rows, we have only λ = 0 as 0 0
.
1 0 9/11 2017-6-6 UW Common Math 308 Section 6.1 12.1/1 points | Previous AnswersHoltLinAlg1 6.1.045. Determine if the statement is true or false, and justify your answer.
If 0 is the only eigenvalue of A, then A must be the zero matrix.
True. Since 0 is the only eigenvalue, there exists a nonzero vector x such that Ax = 0, and
thus A = 0.
True. Since 0 is the only eigenvalue, Ax = 0 for each vector x, and thus A = 0. 00
False. Consider .
01
False. The zero matrix has the eigenvalue 1.
00 False. Consider .
10 Solution or Explanation
0 0
False, has only the eigenvalue 0.
1 0 13.1/1 points | Previous AnswersHoltLinAlg1 6.1.046. Determine if the statement is true or false, and justify your answer.
The product of the eigenvalues (counting multiplicities) of A is equal to the constant term of
the characteristic polynomial of A. True. Since det(A − λIn) = (λ1 − λ)(λ2 − λ) (λn − λ), we conclude that the product of
the eigenvalues is equal to the constant term of the characteristic polynomial.
True. Since det(A − λ2In) = (λ1 − λ2)(λ2 − λ2) (λn − λ2), we conclude that the product
of the eigenvalues is equal to the constant term of the characteristic polynomial. False. Since det(A − λ2In) = (λ1 − λ2)(λ2 − λ2) (λn − λ2), we conclude that the product
of the eigenvalues is equal to the leading coefficient of the characteristic polynomial.
False. Since det(A − λIn) = (λ1 − λ)(λ2 − λ) (λn − λ), we conclude that the product of
the eigenvalues is equal to the leading coefficient of the characteristic polynomial.
10
False. Consider .
01 Solution or Explanation
True. Since det(A − λIn) = (λ1 − λ)(λ2 − λ) (λn − λ), we conclude that the product of the
eigenvalues is equal to the constant term of the characteristic polynomial. 10/11 2017-6-6 UW Common Math 308 Section 6.1 14.4/4 points | Previous AnswersHoltLinAlg1 6.1.047. Suppose that A is a square matrix with characteristic polynomial (λ − 4)3(λ − 2)2(λ + 1).
(a) What are the dimensions of A? (Give n such that the dimensions are n × n.) n = 6 6 (b) What are the eigenvalues of A? (Enter your answers as a commaseparated list.) λ = $$4, 2, −1 (c) Is A invertible? Yes
No (d) What is the largest possible dimension for an eigenspace of A? 3 3 Solution or Explanation
(a) A is 6 × 6. (b) λ = 4, λ = 2, and λ = −1. (c) A is invertible, since 0 is not an eigenvalue. (d) The largest possible dimension of an eigenspace is 3, corresponding to λ = 4. 11/11 ...

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