Math 121: Linear Algebra and Applications
Prof. Lydia Bieri
Solution Set 2
Posted: Fri. Oct. 17, 2007
Written by: Luca Candelori
Exercise 1
(1.4/12)
.
Recall that for ’if and only if’ statements we need to prove both directions: assuming the
statement to the left prove the one to the right, and viceversa.
In our case, suppose that
W
is a subspace of
V
. For any set
S
, it is always true that
S
⊆
span(
S
),
so obviously
W
⊆
span(
W
). Take now a basis for
W
, call it
w
1
,...,w
k
. Then a typical element of
span(
W
) is given by
w
=
k
X
i
=0
a
i
w
i
but this is in
W
, since
W
is closed under addition and scalar multiplication. Hence
W
= span(
W
).
Conversely, suppose that span(
W
) =
W
. We know that 0
∈
span(
W
), since 0
·
w
is in span(
W
).
Therefore 0
∈
W
, since they are equal. Let now
v,w
∈
W
. Then
v
+
w
∈
span(
W
) by deﬁnition,
and so is
c
·
v
for any
c
and
v
. Therefore
W
is a subspace.
Exercise 2.
(a) (1.5/4) Suppose the vectors
e
1
,...,e
n
are linearly dependent. Then we can ﬁnd scalars
a
1
,...,a
n
, not all zero, such that
a
1
1
0
.
.
.
0
0
+
a
2
0
1
.
.
.
0
0
+
...
+
a
n
0
0
.
.
.
0
1
=
0
0
.
.
.
0
0
but now, looking at each component, we get that
a
1
=
a
2
=
...
=
a
n
= 0, which is a
contradiction. Therefore
e
1
,...,e
n
are linearly independent.
(b) (1.5/5) Again, we apply the same method. Suppose there exist scalars
a
1
,...,a
n
, not all zero,
such that
f
(
x
) =
a
1
+
a
2
x
+
...
+
a
n
x
n
= 0
Then the polynomial
f
is of degree
k
, for some 0
≤
k
≤
n
, and so it should have at most
k
roots. But this polynomials is zero everywhere, which implies that
a
1
=
a
2
=
...
=
a
n
= 0.
Exercise 3.
1