Download Solutions #5

APSC 174J, Solutions #5
Posted: March 30, 2016
Section 8, Problem 1(b).
Let ~v = (0, 0, 1), then easy to see ~v cannot be written as a linear combination of
~v2 = (1, 1, 0). So {~v2 } is not a generating set for R3 .
Section 8, Problem 1(g).
Let ~v = (1, 1, 1), then ~v cannot be written as a linear combination of ~v2 = (1, 1, 0) and
~v3 = (1, 2, 1). To see this, we solve the linear system: x1~v2 + x2~v3 = ~v :

1 1 1


1 1 1


1 1 1







 1 2 1 → 0 1 0 → 0 1 0 






0 1 1
0 1 1
0 0 1
From the last row of the row-echelon from, we see there are no solutions for x1 , x2 ,
which means ~v is not in the linear span of {~v2 , ~v3 }.
Hence {~v2 , ~v3 } is not a generating set for R3 .
Note: Since the dimension of R3 is 3, we know any generating set for R3 should have
at least three vectors. So it is easy to see that {~v2 } and {~v2 , ~v3 } are not generating sets
for R3 .
Section 8, Problem 1(k).
To show {~v1 , ~v3 , ~v4 } is a generating set for R3 , we need to show that any vector
~v = (b1 , b2 , b3 ) in R3 can be written as a linear combination of ~v1 , ~v3 , ~v4 . That is, the
linear system: x1~v1 + x2~v3 + x3~v4 = ~v always have a solution.

1 1 0 b1


1 1 0
b1





 0 2 0 b2  →  0 2 0

b
2




0 1 3 b3
0 0 3 b3 − b2 /2
From the row-echelon form, we know the linear system always has a solution no matter
what the values for b1 , b2 , b3 are. So {~v1 , ~v3 , ~v4 } is a generating set for R3 .
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Section 8, Problem 3(a).
Let ~v = (1, −3, 0, 0), then ~v is in W , since the entries of ~v satisfy the conditions of W :
3(1) + (−3) − 0 = 0 and 0 − 2 · 0 = 0
Since ~v = (1, −3, 0, 0) cannot be written as a linear combination of ~v1 = (1, 0, 6, 3), we
know {~v1 } does not generate W , i.e. it is not a generating set for W .
Section 8, Problem 3(f ).
To show {~v2 , ~v3 } is a generating set for W , we need to show that the vector equation:
α1 (0, 1, 2, 1) + α2 (1, 1, 8, 4) = (x, y, z, w)
has a solution for any (x, y, z, w) in W .
Claim: α1 = y − x, α2 = x is a solution for the equation.
In fact, if (x, y, z, w) is in W , then 3x + y − w = 0 and z − 2w = 0. Hence if we express
z, w using x, y, we get z = 2w = 6x + 2y, w = 3x + y. Thus
α1 (0, 1, 2, 1) + α2 (1, 1, 8, 4) = (y − x)(0, 1, 2, 1) + x(1, 1, 8, 4) = (x, y, 6x + 2y, 3x + y)
i.e. α1 = y − x, α2 = x is a solution for the vector equation, so any vector in W can be
written as a linear combination of ~v2 , ~v3 . That is, {~v2 , ~v3 } is a generating set for W .
Section 8, Problem 4(e).
Now we know the dimension of R2 is 2, so any basis of R2 must have exactly two
vectors. Since B1 = (~v1 , ~v2 ) contains two vectors, to show it is a basis, we only need to
show it is linearly independent.
Let α1 (1, 1) + α2 (−1, 1) = (0, 0):


1 −1 0
1
1
0


→
1 −1 0
0
2
0


the only solution is α1 = α2 = 0, so {~v1 , ~v2 } is linearly independent. Hence it is a basis
for R2 .
Section 8, Problem 4(l).
Since any basis of R2 must have exactly two vectors, we know immediately that
(~v1 , ~v2 , ~v4 ) is not a basis for R2 .
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Section 8, Problem 5(b).
Let ~v = (1, 0, 0, 1), then ~v is in W , since the entries of ~v satisfy the conditions of W :
1 + 0 + 0 − 1 = 0 and 0 − 0 = 0
Since ~v = (1, 0, 0, 1) cannot be written as a linear combination of ~v2 = (0, 1, 1, 2), we
know {~v2 } does not generate W , i.e. it is not a generating set for W .
Section 8, Problem 5(g).
To show B3 = (~v3 , ~v1 ) is a basis of W , we need to show {~v1 , ~v3 } is a generating set for
W and also {~v1 , ~v3 } is linearly independent.
1. To show {~v1 , ~v3 } is a generating set for W , we need to show that the vector
equation:
α1 (1, 0, 0, 1) + α2 (1, 1, 1, 3) = (x, y, z, w)
has a solution for any (x, y, z, w) in W .
Claim: α1 = x − y, α2 = y is a solution for the equation.
In fact, if (x, y, z, w) is in W , then x + y + z − w = 0 and y − z = 0. Hence if we
express z, w using x, y, we get z = y, w = x + 2y. Thus
α1 (1, 0, 0, 1, ) + α2 (1, 1, 1, 3) = (x − y)(1, 0, 0, 1, ) + y(1, 1, 1, 3) = (x, y, y, x + 2y)
i.e. α1 = x − y, α2 = y is a solution for the vector equation, so any vector in W
can be written as a linear combination of ~v1 , ~v3 . That is, {~v1 , ~v3 } is a generating
set for W .
2. To show {~v1 , ~v3 } is linearly independent, we solve the vector equation:
α1 (1, 0, 0, 1) + α2 (1, 1, 1, 3) = (0, 0, 0, 0)
Easy to see the solution is α1 = α2 = 0, which implies that {~v1 , ~v3 } is linearly
independent.
Combining 1. and 2. we conclude that B3 = (~v3 , ~v1 ) is a basis of W .
Section 8, Problem 5(o).
1. The component vector of ~v2 with respect to the basis B1 = (~v1 , ~v2 ) is the pair of
numbers (α1 , α2 ) so that α1~v1 + α2~v2 = ~v2 .
α1 (1, 0, 0, 1) + α2 (0, 1, 1, 2) = (0, 1, 1, 2)
The solution is α1 = 0, α2 = 1. So the component vector of ~v2 with respect to the
basis B1 is (0, 1).
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2. The component vector of ~v2 with respect to the basis B2 = (~v1 , ~v3 ) is the pair of
numbers (α1 , α2 ) so that α1~v1 + α2~v3 = ~v2 .
α1 (1, 0, 0, 1) + α2 (1, 1, 1, 3) = (0, 1, 1, 2)
The solution is α1 = −1, α2 = 1. So the component vector of ~v2 with respect to
the basis B1 is (−1, 1).
3. The component vector of ~v2 with respect to the basis B3 = (~v3 , ~v1 ) is the pair of
numbers (α1 , α2 ) so that α1~v3 + α2~v1 = ~v2 .
α1 (1, 1, 1, 3) + α2 (1, 0, 0, 1) = (0, 1, 1, 2)
The solution is α1 = 1, α2 = −1. So the component vector of ~v2 with respect to
the basis B1 is (1, −1).
4. The component vector of ~v2 with respect to the basis B4 = (~v1 , ~v4 ) is the pair of
numbers (α1 , α2 ) so that α1~v1 + α2~v4 = ~v2 .
α1 (1, 0, 0, 1) + α2 (1, −1, −1, −1) = (0, 1, 1, 2)
The solution is α1 = 1, α2 = −1. So the component vector of ~v2 with respect to
the basis B1 is (1, −1).