Homework # 7 Solutions
Math 152, Fall 2014
Instructor: Dr. Doreen De Leon
p. 214-5: 26, 32, 34 (Section 4.3)
26. In the vector space of all real-valued functions, find a basis for the subspace spanned by
{sin t, sin 2t, sin t cos t}.
Solution: First, we know that the functions are linearly dependent, because
sin t cos t =
1
sin 2t.
2
So, 0 sin t + 1 sin 2t − 2 sin t cos t = 0 is a linear dependence relation among the functions.
We can find a basis for S = Span{sin t, sin 2t, sin t cos t} by taking two of the functions
that are linearly independent. As an example, we choose {sin t, sin 2t}.
We can show that {sin t, sin 2t} is linearly independent using the Wronskian:
sin t sin 2t = 2 cos 2t sin t − cos t sin 2t.
W (t) = W [sin t, sin 2t](t) = cos t 2 cos 2t
The set is linearly independent if we can find a value of t such that W (t) 6= 0. If we let
π
t = , for example, we have
4
π π π π
π
= 2 cos 2
sin − cos sin 2
W
4
4
4
4
4
π
π
π
π
= 2 cos sin − cos sin
4
4
2
√ 2
2
=−
.
2
√
π 2
Since W
= −
6= 0, {sin t, sin 2t} is linearly independent and is a basis for the
4
2
vector space spanned by {sin t, sin 2t, sin t cos t} (by the Spanning Set Theorem).
32. Suppose that T is a one-to-one transformation, so that an equatioN T (u) = T (v) always
implies u = v. Show that if the set of images {T (v1 , . . . , T (vp )} is linearly dependent,
then {v1 , . . . , vp } is linearly dependent.
Solution: Since {T (v1 , . . . , T (vp )} is linearly dependent, there exist constants c1 , . . . , cp
not all zero such that
c1 T (v1 ) + · · · + cp T (vp ) = 0.
1
Since T is a linear transformation, this means that
T (c1 v1 + · · · + cp vp ) = 0.
Since T is a linear transformation, we know that T (0) = 0, and since T is one-to-one, we
know that it must be true that
c1 v1 + · · · + cp vp = 0.
Since c1 , . . . , cp are not all zero, we have a linear dependence relation among the vectors
v1 , . . . , vp . Therefore, {v1 , . . . , vp } is linearly dependent.
34. Consider the polynomials p1 (t) = 1 + t, p2 (t) = 1 − t, and p3 (t) = 2 (for all t). By
inspection, write a linear dependence relation among p1 , p2 , and p3 . Then, find a basis
for Span{p1 , p2 , p3 }.
Solution: We can see that 1(t + 1) + 1(1 − t) + (−1)2 = 0. Therefore, {p1 , p2 , p3 } is
linearly dependent.
Claim. {p1 , p3 } is a basis for Span{p1 , p2 , p3 }.
Proof. We just need to show that {p1 , p3 } is linearly independent. To do this, we need to
show that c1 p1 + c2 p3 = 0 has only the trivial solution.
c1 (1 + t) + c2 2 = 0
c1 t + (c1 + 2c2 ) = 0.
Equate coefficients
c1 = 0
c2 + 2c2 = 0 =⇒ c2 = 0.
Since c1 = c2 = 0 is the only solution, {p1 , p3 } is linearly dependent. Therefore, by the
Spanning Set Theorem, it is a basis for Span{p1 , p2 , p3 }.
Note that we could also have used the Wronskian to show that {p1 , p3 } is linearly independent, since
1 + t 2
= −2 6= 0.
W [p1 , p3 ](t) = 1
0
Note: We can also show that both {p1 , p2 } and {p2 , p3 } are also bases for Span{p1 , p2 , p3 }.
p. 223: 22, 26, 32 (Section 4.3)
22. Let B = {b1 , . . . , bn } be a basis for Rn . Produce a description of an n × n matrix A that
implements the coordinate mapping x 7→ [x]B .
Solution: Let PB = b1 b2 · · · bn . Then PB [x]B = x and [x]B = PB−1 x. Therefore,
the matrix A = PB−1 is the desired matrix.
2
26. Given vectors u1 , . . . , up and w in V , show that w is a linear comibination of u1 , . . . , up
if and only if [w]B is a linear combination of the coordinate vectors [u1 ]B , . . . , [up ]B .
Solution:
=⇒ If w = c1 u1 + · · · + cp up for some c1 , . . . , cp , then
[w]B = [c1 u1 + · · · + cp up ]B
= [c1 u1 ]B + · · · + [cp up ]B
= c1 [u1 ]B + · · · + cp [up ]B .
Therefore, [w]B is a linear combination of [u1 ]B , . . . , [up ]B .
⇐= If [w]B = c1 [u1 ]B + · · · + cp [up ]B , then
[w]B = c1 [u1 ]B + · · · + cp [up ]B
= [c1 u1 ]B + · · · + [cp up ]B
= [c1 u1 + · · · + cp up ]B .
Then, since x 7→ [x]B is one-to-one,
w = c1 u1 + · · · + cp up .
So, w is a linear combination of u1 , . . . , up .
32. Let p1 (t) = 1 + t2 , p2 (t) = t − 3t2 , p3 (t) = 1 + t − 3t2 .
(a) Use coordinate vectors to show that these polynomials form a basis for P2 .
Solution: Consider the standard basis B = {1, t, t2 } of P2 . Then,
 
 
 
1
0
1





[p1 ]B = 0 , [p2 ]B = 1 , [p3 ]B = 1  .
1
−3
−3
We need to show that {p1 , p2 , p3 } is linearly independent and spans P2 , which is
equivalent to showing that {[p1 ]B , [p2 ]B , [p3 ]B } is linearly independent and spans R3 ,
because P2 is isomorphic to R3 . Let


1 0
1
1 .
A = 0 1
1 −3 −3
The columns of A are linearly independent and span R3 if det A 6= 0.
1 0
1
1 |A| = 0 1
1 −3 −3
1 1 1+1 1
1+3 0
= 1(−1) + 1(−1) −3 −3
1 −3
= 1(0) + 1(−1) = −1 6= 0.
3
Since det A 6= 0, the columns of A, [p1 ]B , [p2 ]B , [p3 ]B span R3 and are linearly independent. Therefore, {p1 , p2 , p3 } spans P2 and is linearly independent. =⇒ {p1 , p2 , p3 }
is a basis for P2 .
 
−1
(b) Consider the basis B = {p1 , p2 , p3 } for P2 . Find q in P2 , given that [q]B =  1 .
2
Solution:
q = −1(p1 ) + 1(p2 ) + 2(p3 )
= −1(1 + t2 ) + 1(t − 3t2 ) + 2(1 + t − 3t2 )
= −1 − t2 + t − 3t2 + 2 + 2t − 6t2 .
Therefore, q(t) = 1 + 3t − 10t2 .
4