MATH 115 SOLUTION SET 3
§
1.6:
10a, b. I’m sure you can do these.
..
11. Suppose
{
u,v
}
is a basis for
V
. We claim that the set
{
u
+
v,au
}
is also
a basis. We must show it is linearly independent and generates
V
. Suppose that
c
1
(
u
+
v
) +
c
2
(
au
) = 0, so that (
c
1
+
c
2
a
)
u
+
c
2
v
= 0. Since
{
u,v
}
is linearly
independent, we must have
c
1
+
c
2
a
= 0 and
c
2
= 0. This implies that
c
1
=
c
2
= 0, so
that
{
u
+
v,au
}
is linearly independent. By must it generate
V
? Here’s a shortcut:
Let
W
= Span
{
u
+
v,au
}
. Then
W
is twodimensional, and it’s a subspace of
V
,
which is also twodimensional, hence
W
=
V
.
The case of
{
au,bv
}
is similar.
12. First we show that
{
u
+
v
+
w,v
+
w,w
}
is linearly independent. Suppose
a
(
u
+
v
+
w
) +
b
(
v
+
w
) +
cw
= 0. Then
au
+ (
a
+
b
)
v
+ (
a
+
b
+
c
)
w
= 0. We get
a
=
a
+
b
=
a
+
b
+
c
= 0 because
{
u,v,w
}
is linearly independent. Solving, we get
a
=
b
=
c
= 0.
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 Spring '10
 SHALOM
 Math, Linear Algebra, Algebra, Vector Space, basis

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