Linear Algebra Exercises-n-Answers.pdf

Answers to Exercises 55
Two.III.1.20
There are many bases. This is a natural one.
( ) ( ) ( ) ( )
1 0 0 1 0 0 0 0
〈 , , , 〉
0 0 0 0 1 0 0 1
Two.III.1.21 For each item, many answers are possible.
(a) One way to proceed is to parametrize by expressing the a 2 as a combination of the other two
a 2 = 2a 1 + a 0 . Then a 2 x 2 + a 1 x + a 0 is (2a 1 + a 0 )x 2 + a 1 x + a 0 and
{(2a 1 + a 0 )x 2 + a 1 x + a 0
∣ ∣ a 1 , a 0 ∈ R} = {a 1 · (2x 2 + x) + a 0 · (x 2 + 1) ∣ ∣ a 1 , a 0 ∈ R}
suggests 〈2x 2 +x, x 2 +1〉. This only shows that it spans, but checking that it is linearly independent
is routine.
(b) Paramatrize { ( a b c ) ∣ a + b = 0} to get { ( −b b c ) ∣ b, c ∈ R}, which suggests using the
sequence 〈 ( −1 1 0 ) , ( 0 0 1 ) 〉. We’ve shown that it spans, and checking that it is linearly
independent is easy.
(c) Rewriting
suggests this for the basis.
( )
a b ∣∣
{ a, b ∈ R} = {a ·
0 2b
( )
1 0
+ b ·
0 0
( ) ( )
1 0 0 1
〈 , 〉
0 0 0 2
( )
0 1 ∣∣
a, b ∈ R}
0 2
Two.III.1.22 We will show that the second is a basis; the first is similar. We will show this straight
from the definition of a basis, because this example appears before Theorem 1.12.
To see that it is linearly independent, we set up c 1 · (cos θ − sin θ) + c 2 · (2 cos θ + 3 sin θ) =
0 cos θ + 0 sin θ. Taking θ = 0 and θ = π/2 gives this system
c 1 · 1 + c 2 · 2 = 0
c 1 · (−1) + c 2 · 3 = 0
ρ 1 +ρ 2
−→
c 1 + 2c 2 = 0
+ 5c 2 = 0
which shows that c 1 = 0 and c 2 = 0.
The calculation for span is also easy; for any x, y ∈ R 4 , we have that c 1 ·(cos θ −sin θ)+c 2 ·(2 cos θ +
3 sin θ) = x cos θ + y sin θ gives that c 2 = x/5 + y/5 and that c 1 = 3x/5 − 2y/5, and so the span is the
entire space.
Two.III.1.23 (a) Asking which a 0 + a 1 x + a 2 x 2 can be expressed as c 1 · (1 + x) + c 2 · (1 + 2x) gives
rise to three linear equations, describing the coefficients of x 2 , x, and the constants.
c 1 + c 2 = a 0
c 1 + 2c 2 = a 1
0 = a 2
Gauss’ method with back-substitution shows, provided that a 2 = 0, that c 2 = −a 0 + a 1 and c 1 =
2a 0 − a 1 . Thus, with a 2 = 0, that we can compute appropriate c 1 and c 2 for any a 0 and a 1 .
So the span is the entire set of linear polynomials {a 0 + a 1 x ∣ a 0 , a 1 ∈ R}. Paramatrizing that
set {a 0 · 1 + a 1 · x ∣ a 0 , a 1 ∈ R} suggests a basis 〈1, x〉 (we’ve shown that it spans; checking linear
independence is easy).
(b) With
a 0 + a 1 x + a 2 x 2 = c 1 · (2 − 2x) + c 2 · (3 + 4x 2 ) = (2c 1 + 3c 2 ) + (−2c 1 )x + (4c 2 )x 2
we get this system.
2c 1 + 3c 2 = a 0
2c 1 + 3c 2 = a 0
ρ 1 +ρ 2 (−4/3)ρ 2 +ρ 3
−2c 1 = a 1 −→ −→ 3c 2 = a 0 + a 1
4c 2 = a 2 0 = (−4/3)a 0 − (4/3)a 1 + a 2
Thus, the only quadratic polynomials a 0 + a 1 x + a 2 x 2 with associated c’s are the ones such that
0 = (−4/3)a 0 − (4/3)a 1 + a 2 . Hence the span is {(−a 1 + (3/4)a 2 ) + a 1 x + a 2 x 2 ∣ ∣ a 1 , a 2 ∈ R}. Paramatrizing
gives {a 1 · (−1 + x) + a 2 · ((3/4) + x 2 ) ∣ ∣ a 1 , a 2 ∈ R}, which suggests 〈−1 + x, (3/4) + x 2 〉
(checking that it is linearly independent is routine).
Two.III.1.24 (a) The subspace is {a 0 + a 1 x + a 2 x 2 + a 3 x ∣ 3 a 0 + 7a 1 + 49a 2 + 343a 3 = 0}. Rewriting
a 0 = −7a 1 − 49a 2 − 343a 3 gives {(−7a 1 − 49a 2 − 343a 3 ) + a 1 x + a 2 x 2 + a 3 x ∣ 3 a 1 , a 2 , a 3 ∈ R},
which, on breaking out the parameters, suggests 〈−7 + x, −49 + x 2 , −343 + x 3 〉 for the basis (it is
easily verified).