MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
#1. Page 242; Exercise 0 1 (a) 2 , 1 0 −1 1 2 (b) 1 , 3 −1 4 −1 3 (c) 2 , 2 1 2 0 1 (d) 0 , 2 −1 0
2. Which of the following sets of vectors are bases for R3 ? 4 0 , 1 , 1 −1 −1 0 , 1 0 3 0 , 4 , 1 1 0
Solution: (a) Not a basis. The vector space R3 has dimension 3. According to Corollary 4.5, any subset of 2 < 3 vectors cannot span R3 . Hence this subset cannot be a basis. (b) Not a basis. According to Corollary 4.4, any subset of 4 > 3 vectors must be linearly dependent. Hence this subset cannot be a basis. (c) A basis. Since this subset has 3 vectors, if we show that it is linearly independent then we may conclude it is a basis by Theorem 4.12. Indeed, consider the equation −1 0 0 3 a1 − a2 = 0 3 2 a1 + 2 a2 + a3 = 0 =⇒ a1 2 + a2 2 + a3 1 = 0 1 0 0 2 a1 + a2 = 0 2 We find the following augmented matrix and its reduced row echelon form: 3 −1 0 0 1 0 0 0 a1 2 0 1 0 0 2 1 0 =⇒ =⇒ 2 1 0 0 0 0 1 0
a2 a3
= 0 = 0 = 0
Hence the only solution is a1 = a2 = a3 = 0. This shows that the subset is linearly independent, and so it must be a basis for R3 . (d) Not a basis. According to Corollary 4.4, any subset of 4 > 3 vectors must be linearly dependent. Hence this subset cannot be a basis.
#2. Page 242; Exercise 3. Which of the following sets of 1 0 0 1 , 0 1 0 0 , 1 1 1 1 (a) 1 −1 0 2 , 3 −1 2 1 , 1 0 0 (b) −2 4 6 4 , 0 1 2 0 , −1 2 3 (c) 1
vectors are bases for R4 ? , 0 1 1 1 1 2 , −3 2 5 6 , −2
−1
0
4
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MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
(d)
0
0
1
1 , −1
1
1
2
, 1
1
0
0
, 2
1
2
1
Solution: (a) A basis. The vector space R4 has dimension 4. Since this subset has 4 vectors, if we show that it is linearly independent then we may conclude it is a basis by Theorem 4.12. Indeed, consider the equation a1 1 0 0 1 + a2 0 1 0 0 + a3 1 1 1 1 + a4 0 1 1 1 = 0 0 0 0 We may express this as the linear system a1 a2 a1
+ a3 + a3 a3 + a3
+ + +
a4 a4 a4
= 0 = 0 = 0 = 0
1 0 0 1
=⇒
0 1 0 0
1 1 1 1
0 1 1 1
0 0 0 0
=⇒
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
0 0 0 0
Hence the only solution is a1 = a2 = a3 = a4 = 0. This shows that the subset is linearly independent, and so it must be a basis for R4 . (b) Not a basis. According to Corollary 4.5, any subset of 3 < 4 vectors cannot span R4 . Hence this subset cannot be a basis. (c) Not a basis. According to Corollary 4.4, any subset of 5 > 4 vectors must be linearly dependent. Hence this subset cannot be a basis. (d) A basis. Since this subset has 4 vectors, if we show that it is linearly independent then we may conclude it is a basis by Theorem 4.12. Indeed, consider the equation a1 0 0 1 1 + a2 −1 1 1 2 + a3 1 1 0 0 + a4 2 1 2 1 = 0 0 0 0 We may express this as the linear system −
a1 a1
a2 a2 + a2 + 2 a2
+ a3 + a3
+ 2 a4 + a4 + 2 a4 + a4
= 0 = 0 = 0 = 0
=⇒
0 0 1 1
−1 1 2 0 1 1 1 0 1 0 2 0 2 0 1 0
=⇒
1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1
0 0 0 0
Hence the only solution is a1 = a2 = a3 = a4 = 0. This shows that the subset is linearly independent, and so it must be a basis for R4 .
3 #3. Page 242; Exercise 8. In Exercises 7 and 8, determine which of the given subsets form a basis for R . 2 Express the vector 1 as a linear combination of the vectors in each subset that is a basis. 3 1 1 1 2 (a) 1 , 2 , 1 , 5 3 1 4 1 2 3 1 (b) 1 , 2 , 4 2 0 −1
Solution: (a) Not a basis. The vector space R3 has dimension 3. According to Corollary 4.4, any subset of 4 > 3 vectors must be linearly dependent. Hence this subset cannot be a basis.
MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
3
(b) A basis. Since this subset has 3 vectors, if we show that it spans R3 then basis by Theorem 4.12. Indeed, consider the equation 1 2 3 a a1 + 2 a2 a1 + 2 a2 a1 1 + a2 2 + a3 4 = b =⇒ 2 0 −1 c 2 a1 We find the following augmented matrix and its reduced 1 2 3 a 4 b =⇒ 1 2 2
0
−1
c
row echelon form: −a+b+c 1 0 0 2 0 1 0 9 a−7 b−c 4 0 0 1 −a + b
we may conclude it is a + 3 a3 + 4 a3 − a3
= = =
a b c
This shows that the subset is is a spanning set, and so it must be a basis for R3 . When a = 2, b = 1, and c = 3 we find the values a1 = 1, a2 = 2, and a3 = −1. Hence we have the linear combination 2 1 2 3 1 = (1) 1 + (2) 2 + (−1) 4 3 2 0 −1
#4. Page 242; Exercise 11. Find a basis for the subspace W of R3 spanned by 3 11 7 1 2 , 2 , 10 , 6 . 2 1 7 4 What is the dimension of W ? Solution: We determine which vectors can be expressed as linear combinations of the others. consider the equation 1 3 11 7 0 a1 + 3 a2 + 11 a3 + a1 2 + a2 2 + a3 10 + a4 6 = 0 =⇒ 2 a1 + 2 a2 + 10 a3 + 2 1 7 4 0 2 a1 + a2 + 7 a3 + We find the following augmented matrix and its reduced row echelon 1 3 11 7 0 1 0 2 2 10 6 0 0 1 =⇒ 2 1 7 4 0 0 0
To this end, 7 a4 6 a4 4 a4
= 0 = 0 = 0
form: 2 1 3 2 0 0
0 0 . 0
The variables a1 and a2 are the pivot variables, so the vectors can all be expressed as linear combinations of the first two vectors. Hence a basis for W is 3 1 2 , 2 =⇒ dim W = 2. 2 1
#5. Page 243; Exercise 14. Let 1 S= 0
0 1
0 , 1
1 0
Find a basis for the subspace W = span S of M22 .
1 , 1
1 1
−1 , 1
1 −1
.
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MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
Solution: We determine which vectors can be expressed as linear combinations of the others. To this end, consider the equation 1 0 0 1 1 1 −1 1 0 0 a1 + a2 + a3 + a4 = . 0 1 1 0 1 1 1 −1 0 0 We may express this as the linear system a1 a2 a2 a1
+ + + +
a3 a3 a3 a3
− + + −
a4 a4 a4 a4
= 0 = 0 = 0 = 0
=⇒
1 0 0 1
0 1 1 0
1 1 1 1
0 0 0 0
−1 1 1 −1
1 0 1 0 1 1 0 0 0 0 0 0
=⇒
−1 1 0 0
0 0 0 0
The variables a1 and a2 are the pivot variables, so the matrices can all be expressed as linear combinations of the first two matrices. Hence a basis for W is 1 0 0 1 , 0 1 1 0
#6. Page 243; Exercise 15. Find all values of a for which 2 a 0 1 , 0 a 2 , 1
0
1
is a basis for R3 . Solution: It suffices to find all values of a for which this subset is a spanning set for R3 . To this end, consider the equation a1 a2 0 1 + a2 0 a 2 + a3 1 0 1 = x y z . We may express this as the linear system a2 a1 a1
+
a a2 2 a2
+
a3
+
a3
= x = y = z
a2 0 1
=⇒
The reduced row echelon form for the augmented matrix is a x+2 y−a z 1 0 0 a3 −a y 0 1 0 a y+a2 z 0 0 1 −x−2a2a−1
0 1 a 0 2 1
x y . z
.
Hence the subset is a spanning set precisely when a3 6= a, i.e., a 6= −1, 0, 1.
#7. Page 243; Exercise 16. Find a basis for the subspace W of M33 consisting of all symmetric matrices. Solution: The subspace of interest is W = A ∈ M33 AT = A . If we write a11 a12 a13 a11 a21 A = a21 a22 a23 =⇒ AT = a12 a22 a31 a32 a33 a13 a23
a31 a32 . a33
MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
5
Hence AT = A precisely when a21 = a12 , a31 = a13 , and a32 = a23 . This means a11 a12 a13 1 0 0 0 0 0 0 0 0 A = a12 a22 a23 = a11 0 0 0 + a22 0 1 0 + a33 0 0 0 a13 a23 a33 0 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 0 0 + a12 1 0 0 + a13 0 0 0 + a23 0 0 1 . 0 0 0 1 0 0 0 1 0 Hence a basis for the subspace of 0 0 1 0 0 0 0 0 , 0 1 0 0 0 0 0
all symmetric 0 0 0 0 , 0 0 0 0 0
3 × 3 matrices 0 0 1 0 , 1 0 1 0 0
is 0 0 0 , 0 0 1
0 0 0
1 0 0 , 0 0 0
0 0 1
0 1 0
#8. Page 243; Exercise 17. Find a basis for the subspace of M33 consisting of all diagonal matrices. Solution: 3 × 3 diagonal matrices are in the a11 0 0 A = 0 a22 0 = a11 0 0 a33
form 1 0 0
0 0 0
0 0 0 + a22 0 0 0
0 1 0
0 0 0 + a33 0 0 0
matrices is 0 0 0 1 0 , 0 0 0 0
0 0 0
0 0 1
0 0 0
0 0 . 1
Hence a basis for the subspace of all 1 0 0
diagonal 3 × 3 0 0 0 0 0 , 0 0 0 0
#9. Page 243; Exercise 19(a), (b). R4 . a (a) All vectors of the form b c a (b) All vectors of the form b c
In Exercises 19 and 20, find a basis for the given subspaces of R3 and , where b = a + c , where b = a
Solution: (a) Such matrices are in the form
a 1 0 A = a + c = a 1 + c 1 . c 0 1
Hence a basis for the subspace is 0 1 1 , 1 0 1
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MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
(b) Such matrices are in the form
0 a 1 A = a = a 1 + c 0 . c 0 1 Hence a basis for the subspace is 0 1 1 , 0 0 1
#10.Page 251; Exercise 2. Let
1 A = −2 −1
1 −2 −2 4 . −1 2
(a) Find the set of all solutions to A x = 0. (b) Express each solution as a linear combination of two vectors in R3 . (c) Sketch these vectors in a three-dimensional coordinate system to show that the solution space is a plane through the origin.
Solution: (a) The system A x = row echelon form: 1 1 −2 −2 −2 4 −1 −1 2
0 has the following augmented matrix, for which we compute its reduced 0 0 0
1 1 0 0 0 0
=⇒
−2 0 0 0 0 0
x + y − 2 z = 0.
Hence y = r and z = s are arbitrary, so that the solution is x = −r + 2 s, y = r, and z = s for arbitrary real numbers r and s. (b) We may express this solution in terms of vectors
x −r + 2 s −1 2 = r 1 + s 0 r x= y = z s 0 1 (c) The set of solutions is the plane x + y − 2 z = 0.
#11. Page 252; Exercise 5. In Exercises 3 through 10, find a basis for and and the dimension of the solution space of the given homogeneous system. x1 2 x1 x1
+ 2 x2 + 2 x2
− − +
x3 x3 3 x3
+ 3 x4 + 2 x4 + 3 x4
= 0 = 0 = 0
MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
7
Solution: We find the following augmented matrix and its reduced row echelon form: x1 1 0 0 −1 0 1 2 −1 3 0 x2 8 0 =⇒ =⇒ 0 1 0 2 2 −1 2 0 3 x 3 4 1 0 3 3 0 0 0 0 1 3 x4
r
8 −3 r = −4 r 3
r
where r is an arbitrary variable. Hence a basis for the set of solutions is 1 − 8 3 −4 3 1
=⇒
dimension = 1.
#12. Page 252; Exercise 7. In Exercises 3 through 10, find a basis for and and the dimension of the solution space of the given homogeneous system. x 1 2 1 2 1 1 0 x2 1 2 2 1 2 x3 = 0 . 2 4 3 3 3 0 x4 0 0 1 −1 −1 0 x5
Solution: We find the following augmented matrix and its reduced row echelon 1 2 0 3 1 2 1 2 1 0 0 0 1 −1 1 2 2 1 2 0 =⇒ 0 0 0 2 4 3 3 3 0 0 0 0 1 −1 −1 0 0 0 0 0
form: 0 0 1 0
0 0 . 0 0
Hence x2 = r and x4 = s are arbitrary variables, so that the general solution to the homogeneous system is x1 −2 r − 3 s −2 −3 x2 1 0 r x3 = = r 0 + s 1 . s x4 0 1 s x5 0 0 0 Hence a basis for the set of solutions is −2 −3 1 0 0 , 1 0 1 0 0
=⇒
dimension = 2.
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MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
#13. Page 252; Exercise 12. In Exercises 11 and 12, find a basis for the null space of each given matrix A. 1 −1 2 1 0 2 0 1 −1 3 5 −1 3 0 3 A= 4 −2 5 1 3 1 3 −4 −5 6
Solution: The null space of A is the set of solutions x to the linear system A x augmented matrix for this system and its reduced row echelon form: 1 −1 2 1 0 0 1 0 0 0 2 0 1 0 −3 0 1 −1 3 0 5 −1 0 0 1 −1 3 0 3 0 =⇒ 4 −2 0 0 0 5 1 3 0 0 1 3 −4 −5 6 0 0 0 0 0
= 0. We find the following 0 6 3 0 0
0 0 0 0 0
.
Hence x4 = r and x5 = s are arbitrary variables, so that the general solution to the homogeneous system is 0 x1 0 0 −6 x2 3 r − 6 s 3 x3 = r − 3 s = r 1 + s −3 . 0 x4 1 r 1 0 s x5 Hence a basis for the null space of A is
0 3 1 1 0
,
0 −6 −3 0 1
#14. Lab 6.4 - Page 21; Exercise 1. Let V = R3 . Determine whether the following sets are a basis for V. It may be possible to decide without any computations. Record your response next to each set. 2 0 1 a) S = 2 , 1 , 3 1 1 1 1 0 1 b) S = 1 , 0 , 1 0 1 1 3 3 c) S = 1 , 1 3 2 3 3 3 3 d) S = 1 , 1 , 1 , 1 3 2 1 0
Solution: a) We perform the following commands in Matlab: >> A = [1 2 1; 2 1 1; 0 3 1]’ A =
MA 265 HOMEWORK ASSIGNMENT #8 SOLUTIONS
1 2 1
2 1 1
9
0 3 1
>> rref(A) ans = 1 0 2 0 1 -1 0 0 0 We see that the vectors in S are linearly dependent, so S is not a basis for V. b) We perform the following commands in Matlab: >> A = [1 1 0; 1 0 1; 0 1 1]’ A = 1 1 0
1 0 1
0 1 1
>> rref(A) ans = 1 0 0 0 1 0 0 0 1 We see that the vectors in S are linearly independent, so S is a basis for V. c) We see from (b) that V is a 3-dimensional vector space. Since S has 2 < 3 elements, it cannot be a spanning set for V. Hence S is not a basis for V. d) Since S has 4 > 3 elements, it cannot be linearly independent. Hence S is not a basis for V.
