Math 104
Fall 2009
Homework 1: solution set
We use the notations below to ease readability. Matrices are bold capital, vectors are bold lowercase and
scalars or entries are not bold. For instance,
A
is a matrix and
a
ij
its (
i, j
)th entry. Likewise
x
is a vector
and
x
j
its
j
th component. The linear span generated by a group of vecotors
v
1
,
v
2
, . . . ,
v
n
is denoted by
span(
v
1
,
v
2
, ...,
v
n
).
Problem 1
(a) Since
v
3
=
v
1
+
v
2
, we have
v
3
∈
span(
v
1
,
v
2
). Thus span(
v
1
,
v
2
,
v
3
) = span(
v
1
,
v
2
). Suppose
a, b
∈
R
,
such that
a
v
1
+
b
v
2
= 0. Then we have
a

b
= 0,
b
= 0 and

a
= 0, which implies
a
=
b
= 0. Hence,
v
1
and
v
2
are linearly independent and, therefore,
v
1
,
v
2
is a basis of span(
v
1
,
v
2
) = span(
v
1
,
v
2
,
v
3
). In conclusion
dim(span(
v
1
,
v
2
,
v
3
)) = 2.
(b) Since
v
1
+
v
2

v
3
= 0,
v
1
,
v
2
,
v
3
are linearly dependent, they cannot form a basis.
(c) There are many possible answers here; one of them is the 1
×
3 matrix
A
=
111
.
Of course, using this example we could construct many other examples, for example for any integer
m
we
can consider an
m
by 3 matrix all of whose rows are scalar multiples of (111). Such a matrix also has the