Assignment 7
Math 417 — Winter 2009
Due March 13
§
3.4#10.
The question is whether we can express
x
=

5
1
3
as a linear combination
c
1
v
1
+
c
2
v
2
of
v
1
=

1
0
1
and
v
2
=

2
1
0
.
By inspection we see that, if there is to be
such an expression, the coefficient
c
1
must be 3 in order to make the last component right
(3 =
c
1
·
1 +
c
2
·
0), and similarly
c
2
must be 1 in order to make the middle component
right. These values for the two coefficients happen to make the first component right also
(

5 = 3
·
(

1)+1
·
(

2)), so
x
is in the span of
v
1
, v
2
, and its coordinates are the coefficients
we found, 3 and 1. Its coordinate vector is [
x
]
B
=
3
1
.
§
3.4#12.
The question is whether we can express
x
=
1

2

2
as a linear combination
c
1
v
1
+
c
2
v
2
of
v
1
=
8
4

1
and
v
2
=
5
2

1
.
In contrast to the preceding problem, there
seems to be no way to guess the coefficients
c
i
by inspection (or even to guess whether
such coefficients exist). So we compute instead. We need 8
c
1
+ 5
c
2
= 1 to make the first
component of
c
1
v
1
+
c
2
v
2
match the first component of
x
. We need 4
c
1
+ 2
c
2
=

2 to make
the second components match. And we need

c
1

c
2
=

2 to make the third components
match. So
x
will be in the span of
v
1
and
v
2
if and only if this system of three equations has
a solution for the two unknowns
c
1
and
c
2
. The augmented matrix of the system is
8
5
1
4
2

2

1

1

2
.
To reduce it to rref, I’ll first interchange the last row with the first (indulging my dislike for
fractions); while I’m at it, I’ll also reverse the sign of the top (formerly bottom) row.
1
1
2
4
2

2
8
5
1
→
1
1
2
0

2

10
0

3

15
→
1
1
2
0
1
5
0
1
5
→
1
0

3
0
1
5
0
0
0
.
1
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So there is a solution, namely
c
1
=

3 and
c
2
= 5, which means that
x
is in the span of
v
1
and
v
2
. Its coordinate vector with respect to this basis is therefore [
x
]
B
=

3
5
.
§
3.4#22.
(a)
To use the formula
B
=
S

1
AS
, recall that
S
is the matrix whose columns are the
vectors
v
i
of the basis. In this problem, therefore,
S
=
1

2
2
1
. We need its inverse
S

1
as well, so we compute it by row reduction:
1

2
1
0
2
1
0
1
→
1

2
1
0
0
5

2
1
→
1

2
1
0
0
1

2
/
5
1
/
5
→
1
0
1
/
5
2
/
5
0
1

2
/
5
1
/
5
.
So
B
=
S

1
AS
=
1
/
5
2
/
5

2
/
5
1
/
5

3
4
4
3
1

2
2
1
=
5
0
0

5
.
(b)
The commutative diagram method (as in the book’s Examples 3 and 4) requires us
to take an arbitrary vector expressed as a linear combination
c
1
v
1
+
c
2
v
2
, apply the
linear transformation (represented by)
A
to it, and express the result again as a linear
combination of
v
1
and
v
2
. There are two ways to do this. The more obvious but less
efficient is to use the given
v
1
and
v
2
to compute that
c
1
v
1
+
c
2
v
2
=
c
1

2
c
2
2
c
1
+
c
2
,
and applying
A
yields

3
4
4
3
c
1

2
c
2
2
c
1
+
c
2
=
5
c
1
+ 10
c
2
10
c
1

5
c
2
.
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 Spring '09
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 Math, Linear Algebra, Algebra, Vectors, basis, Linear combination, v2, basis vectors

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