Mathematics 206 Solutions for HWK 15c Section 5 - Wellesley College

Mathematics 206 Solutions for HWK 15c

Section 5.4

Section 5.4 p244 Problem 6. Let V be the space spanned by v1 = cos2 x, v2 = sin2 x, v3-cos 2x.

(a) Show that S = {v1, v2, v3} is not a basis for V .

(b) Find a basis for V .

Solution. (a) S cannot be a basis for V because it is not independent: v3 = v1 - v2.

(b) Discard the vector v3. The new set {v1, v2} is easily seen to be independent (we know from trigonometry that neither of these functions is a scalar multiple of the other). Moreover it has the same span as the original set S. Therefore it is an independent spanning set for V , in other words a basis for V . Other correct answers are possible. In fact, in this problem one could discard any one of the vectors in S and arrive at a basis for V .

Section 5.4 p244 Problem 7bc. Find the coordinate vector of w relative to the basis S = {u1, u2} for R2.

(b) u1 = (2, -4), u2 = (3, 8), w = (1, 1).

(c) u1 = (1, 1), u2 = (0, 2), w = (a, b).

Solution. (b) We need to solve the vector equation (1, 1) = k1(2, -4) + k2(3, 8), in other words solve the system

2k1 + 3k2 = 1

-4k1 + 8k2 = 1

I'll use Cramer's Rule but you might prefer to use some other method. According to Cramer's

Rule,

k1

=

D1 , D

k2

=

D2 D

where

D1 = det

1 1

3 8

,

D2 = det

2 -4

1 1

,

D = det

2 -4

3 8

5

63

This

gives

k1

=

, 28

k2

=

28

=

. 14

Therefore

the

desired

coordinate

vector

is

53 1

(w)S

=

(, 28

) 14

=

(5, 6). 28

(c) This time we need k1 and k2 so that (a, b) = k1(1, 1) + k2(0, 2). Evidently k1 = a. Then we

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A. Sontag April 6, 2002

Math 206 HWK 15b Solns contd 5.4 p244

b-a need b = k1 + 2k2 = a + 2k2, which gives 2k2 = b - a or k2 = 2 . The coordinate vector is

b-a (w)S = (a, 2 ).

Section 5.4 p244 Problem 8a. Find the coordinate vector of v = (2, -1, 3) relative to the basis {v1, v2, v3}, given that v1 = (1, 0, 0), v2 = (2, 2, 0), and v3 = (3, 3, 3).

Solution. We seek a, b, c so that (2, -1, 3) = a(1, 0, 0) + b(2, 2, 0) + c(3, 3, 3). Using the third entry we find 3c = 3 so c = 1. Then the second entry gives 2b + 3 = -1 or 2b = -4 so b = -2. Finally, the first entry then gives us a - 4 + 3 = 2 or a = 3. The desired coordinate vector for v is therefore (3, -2, 1).

Section 5.4 p244 Problem 9b. Find the coordinate vector of p relative to the basis S = {p1, p2, p3} for P2, given that p(x) = 2 - x + x2, p1(x) = 1 + x, p2(x) = 1 + x2, p3(x) = x + x2.

Solution. We need to find a, b, c so that a(1 + x) + b(1 + x2) + c(x + x2) = 2 - x + x2

for every x. Equating coefficients we convert this to the system a+b=2 a + c = -1 b+c=1

Using whatever method you like, solve this system to find a = 0, b = 2, c = -1. Therefore (p)S = (0, 2, -1).

Section 5.4 p244 Problem 10. Find the coordinate vector of A relative to the basis S = {A1, A2, A3, A4}, given that

A=

2 -1

0 3

,

A1 =

-1 0

1 0

,

A2 =

1 0

1 0

,

A3 =

0 1

0 0

,

A4 =

0 0

0 1

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A. Sontag April 6, 2002

Math 206 HWK 15b Solns contd 5.4 p244

Solution. We need scalars a, b, c, d such that A = aA1 + bA2 + cA3 + dA4. Equating individual entries in the respective matrices we find the following system of equations:

-a + b = 2 a+b=0 c = -1

d=3 This gives a = -1, b = 1, c = -1, d = 3 so (A)S = (-1, 1, -1, 3).

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A. Sontag April 6, 2002

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