Solutions to HW #3
Math 67
UC Davis, Fall 2011
1. Solve the following problems in the textbook:
(a) Calculational exercises 1, 2, 3 in Chapter 5.
(b) Proofwriting exercise 4 in Chapter 5.
Solution to calculational exercise 1.
If
a
1
,a
2
,a
3
are scalars such that
a
1
v
1
+
a
2
v
2
+
a
3
v
3
= 0, this gives the homogeneous linear system of equations
a
1
+
a
2
+
a
3
= 0
a
1
+ 2
a
2

a
3
= 0
a
1
+ 3
a
2
+
a
3
= 0
By simplifying the system using Gaussian elimination, one can check that
a
1
=
a
2
=
a
3
= 0 is the unique solution, hence the vectors
v
1
,v
2
,v
3
are linearly independent.
Solution to calculational exercise 2.
(a) Denote
e
1
= (1
,
0
,
0),
e
2
= (0
,
1
,
0),
e
3
= (0
,
0
,
1), the standard basis vectors. Note that
e
1
= (

i
)
v
1
,
e
2
=
v
2

v
1
,
e
3
=

v
3
+
v
1
+
ie
2
=

v
3
+
v
1
+
i
(
v
2

v
1
)
,
i.e.,
e
1
,e
2
,e
3
∈
span(
v
1
,v
2
,v
3
), and therefore also
V
=
C
3
, which is the span of
e
1
,e
2
,e
3
, is contained in span(
v
1
,v
2
,v
3
) (since the span is closed under linear com
binations). We also have the opposite containment relation span(
v
1
,v
2
,v
3
)
⊂
V
, so
combining the two we get the claim that
V
= span(
v
1
,v
2
,v
3
).
(b) We know from (a) above that (
v
1
,v
2
,v
3
) is a spanning set of vectors, and since
it has exactly 3 vectors (the dimension of
V
), it must be linearly independent, since
otherwise (by the “basis reduction theorem”, theorem 5.3.4 on page 55 of the textbook)
we could discard some of the vectors and get a basis with less than 3 elements.