Selected Answers to 3.4 and 4.4 Suggested Problems
3.4
10. (a)
S
is clearly linearly independent, so we need only verify it is in
V
. 0

2+3

1+0 = 0,
so it is.
(b) We need a spanning set for
V
to which we can add
S
, then pare down into a basis.
V
has dimension 4, since determining any four entries ﬁxes the ﬁfth, but determining 3 leaves
you freedom. Some vectors of
V
: (1
,
0
,
0
,
1
,
0), (0
,
1
,
0
,
0
,
1), (1
,
0
,

1
,
0
,
1), (2
,
1
,
0
,
0
,
0). This
is a linearly independent set and hence a basis. If we make a matrix with
S
as the ﬁrst
column (this ensures
S
will be included in our basis) and this basis as the remaining four
columns and rowreduce it, we get
1 0 0

1 0
0 1 0
1
0
0 0 1
1
0
0 0 0
0
1
0 0 0
0
0
.
Since the ﬁrst, second, third, and ﬁfth columns correspond to the ﬁrst four elements of the
standard basis, they correspond to our basis elements. Hence, a basis for
V
which includes
S
would be
{
(0
,
1
,
1
,
1
,
0)
,
(1
,
0
,
0
,
1
,
0)
,
(0
,
1
,
0
,
0
,
1)
,
(2
,
1
,
0
,
0
,
0)