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Linear Algebra — Homework 4
Solutions
1. Write the following vectors as linear combinations of (2, 1, −1), (3, 1, 1), (−4, 0, 6).
(a) (1, 2, 6)
(1, 2, 6) = 1(2, 1, −1) + 1(3, 1, 1) + 1(−4, 0, 6)
(b) (1, 23 , 1)
(1, 23 , 1) = 13 (2, 1, −1) + 31 (3, 1, 1) + 61 (−4, 0, 6)
2. Determine whether each of the following vectors is a linear combination of (1, 1, 1, 0), (1, 0, 0, 1),
(0, 1, 0, 1).
(a) (1, 5, 3, 0)
(1, 5, 3, 0) = 3(1, 1, 1, 0) − 2(1, 0, 0, 1) + 2(0, 1, 0, 1).
(b) (1, 1, 1, −3)
(1, 1, 1, −3) is not a linear combination of these vectors.
3. Write the polynomial 4x3 − 7x2 − 3x + 8 as a linear combination of x + 1, x2 + x, x3 + x2 .
4x3 − 7x2 − 3x + 8 = 8(x + 1) − 11(x2 + x) + 4(x3 + x2 ).
4. (a) Show that {(2, 4, −2), (3, 2, 0), (1, −2, −2)} spans R3 .
To show that {(2, 4, −2), (3, 2, 0), (1, −2, −2)} spans R3 , we will choose a representative
element of R3 and show that it can be written as a linear combination of the three
vectors. Let (x, y, z) ∈ R3 . We attempt to solve the equation
a(2, 4, −2) + b(3, 2, 0) + c(1, −2, −2) = (x, y, z)
(2a + 3b + c, 4a + 2b − 2c, −2a − 2c) = (x, y, z)
So we get the system of equations
2a + 3b + c = x
4a + 2b − 2c = y
−2a
− 2c = z.
Adding the first and third equations, we get
3b − c = x + z.
(1)
Now take 2 times the third equation and add it to the second to get
2b − 6c = y + 2z.
(2)
. Plugging
Take 6 times (1) and subtract (2) to get 16b = 6x − y + 4z. So b = 6x−y+4z
16
18x−3y+12z
− x − z. Then from the first equation in the system we
into (1) we get c =
16
18x−3y+12z
18x−3y+12z
1
get a = 2 x −
−(
− x − z) . Thus for any x, y and z the system has
16
16
3
a solution. Hence the vectors span R .
(b) Show that {(2, 4, −2), (3, 2, 0), (1, −2, 2)} does not span R3 .
We will show that (1, 0, 0) is not in the span of {(2, 4, −2), (3, 2, 0), (1, −2, 2)} and hence
this set does not span R3 . We attempt to solve the equation
a(2, 4, −2) + b(3, 2, 0) + c(1, −2, 2) = (1, 0, 0)
(2a + 3b + c, 4a + 2b − 2c, −2a + 2c) = (1, 0, 0)
So we get the system of equations
2a + 3b + c = 1
4a + 2b − 2c = 0
−2a
+ 2c = 0.
The third equation tells us c = a. Plugging into the second equation we get b = −a.
Plugging these into the first equation gives us 0 = 1. Thus the system has no solution.
5. (a) If possible, give a set with five elements that spans P3 .
{1, x, x2 , x3 , 2x2 − 5x + 9}.
(b) If possible, give a set with four elements that spans P3 .
{1, x, x2 , x3 }.
(c) If possible, give a set with three elements that spans P3 .
Impossible.
6. Let W = {ax3 + 2bx2 − 3(a + b)x + (a − 2b) ∈ P3 | a, b ∈ R}
(a) Write down two different elements of W .
Letting a = 0 and b = 0 we get 0x3 + 0x2 + 0x + 0, which is the 0-polynomial. Letting
a = 1 and b = 0 we get x3 − 3x + 1.
(b) Find an element of P3 that is not an element of W .
We claim that the constant polynomial 1 is not in W . If it were, then we could find a
and b such that ax3 + 2bx2 − 3(a + b)x + (a − 2b) = 1. Setting coefficients on the left
equal to those on the right, we get a = 0 and b = 0, but a − 2b = 1. Thus there are no
such a and b.
(c) Find a finite set of elements of W that spans W . Note: {1, x, x2 , x3 } is wrong.
Rewrite W as follows:
W = {ax3 + 2bx2 − 3(a + b)x + (a − 2b) ∈ P3 | a, b ∈ R}
= {ax3 + 2bx2 − 3ax − 3bx + a − 2b ∈ P3 | a, b ∈ R}
= {a(x3 − 3x + 1) + b(2x2 − 3x − 2) ∈ P3 | a, b ∈ R}
Thus every element of W is a linear combination of x3 − 3x + 1 and 2x2 − 3x − 2. That
is, {x3 − 3x + 1, 2x2 − 3x − 2} spans W . Letting a = 1 and b = 0 demonstrates that
x3 − 3x + 1 ∈ W . Similarly, letting a = 0 and b = 1 demonstrates that 2x2 − 3x − 2 ∈ W .
(d) Show that W is a subspace of P3 .
You could use the subspace theorem, but it’s much more efficient to cite Homework 3,
problem 8. Since W is the span of two vectors in P3 , W is a subspace of P3 .
7. Suppose {~v1 , ~v2 , . . . , ~vk } spans the vector space V , and let w
~ be another vector in V . Prove
that {~v1 , ~v2 , . . . , ~vk , w}
~ also spans V .
Proof: We wish to show that {~v1 , ~v2 , . . . , ~vk , w}
~ spans V . So we must choose a representative
element of V and show that it can be written as a linear combination of those vectors. Let
~x ∈ V . Since {~v1 , ~v2 , . . . , ~vk } spans V , we know we can write
~x = a1~v1 + a2~v2 + · · · + ak~vk .
But then ~x = a1~v1 + a2~v2 + · · · + ak~vk + 0w,
~
since 0w
~ = ~0. Thus ~x is a linear combination of ~v1 , ~v2 , . . . , ~vk , w.
~ Since ~x is representative,
this means {~v1 , ~v2 , . . . , ~vk , w}
~ spans V.