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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.
{u1 , u2 } for R2 .
Find the coordinate vector of w relative to the basis S =
(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,
D1
D2
k1 =
, k2 =
D
D
where
This gives k1 =
·
1
D1 = det
1
¸
·
¸
·
3
2 1
2
, D2 = det
, D = det
8
−4 1
−4
3
8
¸
5
6
3
, k2 =
=
. Therefore the desired coordinate vector is
28
28
14
(w)S = (
5 3
1
, )=
(5, 6).
28 14
28
(c) This time we need k1 and k2 so that (a, b) = k1 (1, 1) + k2 (0, 2). Evidently k1 = a. Then we
Page 1 of 3
A. Sontag
April 6, 2002
Math 206 HWK 15b Solns contd
5.4 p244
need b = k1 + 2k2 = a + 2k2 , which gives 2k2 = b − a or k2 =
(w)S = (a,
b−a
. The coordinate vector is
2
b−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
·
¸
·
¸
·
¸
·
¸
·
¸
2 0
−1 1
1 1
0 0
0 0
A=
, A1 =
, A2 =
, A3 =
, A4 =
−1 3
0 0
0 0
1 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