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MATH 110: LINEAR ALGEBRA
SPRING 2007/08
PROBLEM SET 5 SOLUTIONS
1.
Let
U,W,V
be ﬁnitedimensional vector spaces over
F
. Let
α,β
∈
F
.
(a) Let
T
:
V
→
W
be a linear transformation. Show that
rank(
T
)
≤
dim(
V
)
.
Solution.
Let dim(
V
) =
n
. Suppose dim(im(
T
)) = rank(
T
)
> n
. Then there must
be at least
n
+ 1 vectors
w
1
,...,
w
n
+1
∈
im(
T
) that are linearly independent (eg. choose
the ﬁrst
n
+ 1 vectors in a basis of im(
T
)). Since
w
1
,...,
w
n
+1
are in im(
T
), there exist
v
1
,...,
v
n
+1
∈
V
such that
T
(
v
i
) =
w
i
,
i
= 1
,...,n
+ 1. Since dim(
V
) =
n
,
v
1
,...,
v
n
+1
must be linearly dependent, so there exist
α
1
,...,α
n
+1
∈
F
, not all zero, such that
α
1
v
1
+
···
+
α
n
+1
v
n
+1
=
0
V
.
Hence we have found
α
1
,...,α
n
+1
∈
F
, not all zero, such that
T
(
α
1
v
1
+
···
+
α
n
+1
v
n
+1
) =
T
(
0
V
)
,
ie.
α
1
T
(
v
1
) +
···
+
α
n
+1
T
(
v
n
+1
) =
0
W
,
ie.
α
1
w
1
+
···
+
α
n
+1
w
n
+1
=
0
W
,
which implies that
w
1
,...,
w
n
+1
are linearly dependent — a contradiction. Hence our
original assumption must have been false, ie. we must have dim(im(
T
))
≤
n
.
(b) Let
S
:
U
→
V
and
T
:
V
→
W
be linear transformations. Show that
rank(
T ◦ S
)
≤
rank(
T
)
and
rank(
T ◦ S
)
≤
rank(
S
)
.
Solution.
Recall the following elementary fact from set theory:
A
⊆
B
implies
f
(
A
)
⊆
f
(
B
) for any function
f
. In particular, since
S
(
U
)
⊆
V
, we must have
T
(
S
(
U
))
⊆ T
(
V
),
ie.
T ◦ S
(
U
)
⊆ T
(
V
), ie.
im(
T ◦ S
)
⊆
im(
T
)
(recall that im(
ϕ
) is just another way of writing
ϕ
(
V
) — both denote the range of
ϕ
). Now
both of these are subspaces of
W
and so
dim(im(
T ◦ S
))
≤
dim(im(
T
))
,
ie.
rank(
T ◦ S
)
≤
rank(
T
)
.
For the second part, we deﬁne the function
ϕ
: im(
S
)
→
W
by
ϕ
(
v
) =
T
(
v
) for all
v
∈
im(
S
). Note that
ϕ
is a linear transformation since for all
v
1
,
v
2
∈
im(
S
) and
α,β
∈
F
,
we have
ϕ
(
α
v
1
+
β
v
2
) =
T
(
α
v
1
+
β
v
2
) =
α
T
(
v
1
)+
β
T
(
v
2
) =
αϕ
(
v
1
)+
βϕ
(
v
2
). So applying
part (a) to
ϕ
, we get
rank(
ϕ
)
≤
dim(im(
S
)) = rank(
S
)
.
(1.1)
Date
: May 3, 2008 (Version 1.0).
1
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View Full DocumentBut note that im(
T ◦ S
)
⊆
im(
ϕ
) — since if
w
∈
im(
T ◦ S
), then
w
=
T
(
S
(
u
)) for some
u
∈
U
and therefore
w
=
T
(
v
) =
ϕ
(
v
) where
v
=
S
(
u
)
∈
im(
S
); hence
w
∈
im(
ϕ
). So
dim(im(
T ◦ S
))
≤
dim(im(
ϕ
))
,
ie.
rank(
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Algebra, Vector Space

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