Review Problems for Exam Three
THEORY
1. Use the Law of Cosines to prove that
x
T
y
=

x

y

cos
(
θ
) for
x, y
∈
R
3
.
By the Law of Cosines,
x

y
2
=
x
2
+
y
2

2
x
y cos
(
θ
), where
θ
is the
angle between
x
and
y
. By definition,
x

y
2
=
x

y, x

y
, which equals
x, x
+
y, y

2
x, y
by properties of inner products. Thus

2
x, y
=

2
x
y cos
(
θ
),
so that
x
T
y
=
x, y
=
x
y cos
(
θ
).
2. Let
A
be an
m
×
n
matrix. Identify
Rowspace
(
A
)
⊥
,
Colspace
(
A
)
⊥
and
N
(
A
)
⊥
in terms of the row, column and null spaces of
A
and/or
A
t
.
rs
(
A
)
⊥
=
ns
(
A
),
cs
(
A
)
⊥
=
ns
(
A
T
) and
ns
(
A
)
⊥
=
rs
(
A
)
3. Prove that
Rowspace
(
A
)
⊥
=
Nullspace
(
A
).
Let
A
have rows
A
T
1
, . . . , A
T
m
.
Suppose first that
x
∈
rs
(
A
)
⊥
, so that
y
T
x
= 0 for all
y
∈
rs
(
A
).
Then in
particular
A
T
i
x
= 0 for each
i
.
It follows that
Ax
=
A
T
1
.
.
.
A
T
m
x
=
→
0 , so that
x
∈
ns
(
A
).
Suppose next that
x
∈
ns
(
A
), so that
Ax
= 0, so that
A
i
x
= 0 for each
i
. Then
for any
y
=
c
1
A
1
+
· · ·
+
c
m
A
m
∈
rs
(
A
),
y
T
x
= (
c
1
A
1
+
. . .
+
c
m
A
m
)
T
x
=
c
1
A
T
1
x
+
· · ·
+
c
m
A
T
m
x
=
c
1
0 +
· · ·
+
c
m
0 = 0,
so that
x
∈
rs
(
A
)
⊥
.
4.
(a) Prove that
N
(
A
) =
N
(
A
T
A
) for any
m
×
n
matrix
A
.
(b) Show that
rk
(
A
T
A
) =
rk
(
A
T
). (c) Show that
A
T
A
→
x
=
A
T
→
b
has a solution for any
→
b
∈
R
m
.
(a) Suppose first that
x
∈
ns
(
A
), so that
Ax
= 0. Then (
A
T
A
)
x
=
A
T
(
Ax
) =
A
T
0 = 0, so that
x
∈
ns
(
A
T
A
).
Suppose next that
x
∈
ns
(
A
T
A
), so that
A
T
Ax
= 0. Then
Ax
∈
ns
(
A
T
). Also,
Ax
∈
cs
(
A
).
But
ns
(
A
T
)
⊥
=
rs
(
A
T
) =
cs
(
A
), so that
Ax
∈
ns
(
A
T
)
⊥
.
Thus
Ax
= (
Ax
)
T
Ax
= 0, so that
Ax
= 0. This means that
x
∈
ns
(
A
).
(b) It follows from (a) that
A
and
A
T
A
have the same nullity.
A
T
A
is an
n
×
n
matrix, so that, by the rank plus nullity theorem,
rk
(
A
T
A
) =
n

null
(
A
T
A
) =
n

null
(
A
) =
rk
(
A
).
(c) It follows from (a) that
rs
(
A
T
A
) =
ns
(
A
T
A
)
⊥
=
ns
(
A
)
⊥
=
rs
(
A
),
so that
cs
(
A
T
A
) =
rs
(
A
T
A
) =
rs
(
A
) =
cs
(
A
T
).
Now
A
T
b
∈
cs
(
A
T
), so that
A
T
b
∈
cs
(
A
T
A
), which means that
A
T
b
=
A
T
Ax
for some
x
.
5. Show that for any matrix
A
, the transformation
L
A
is an isomorphism from
the Row Space of
A
onto the Column Space of
A
.
Let
A
be
m
×
n
, so that
L
A
maps
R
n
into
R
m
.
L
A
(
v
) =
Av
is in colspace(A)
for any
v
, so
L
A
maps rowspace(A) into colspace(A).
To show that this map is onto, take any
y
∈
R
(
A
) and let
y
=
Ax
for some
x
∈
R
m
. Now let
x
=
u
+
v
where
u
∈
rowspace
(
A
) and
v
∈
N
(
A
), by Theorem
5.2.3.
Then
Ax
=
Au
+
Av
=
Au
, since
Av
= 0.
Therefore
y
=
Au
with
u
in
rowspace(A).
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To show that this map is onetoone, suppose that
Ax
=
Au
for some
x
and
u
in
rowspace(A). Then
A
(
x

u
) = 0, so that
x

u
∈
N
(
A
). But
x

u
∈
rowspace
(
A
),
which is
N
(
A
)
⊥
, so that
x

u
= 0 and hence
x
=
u
.
6. State the properties of an inner product and show that the standard scalar
product on
R
n
satisfies those properties.
(i)
u, u
≥
0 and
u, u
= 0 if and only if
u
=
→
0 .
(ii)
u, v
=
v, u
(iii)
λu, v
=
λ u, v
for any scalar
λ
.
(iv)
u, v
+
w
=
u, v
+
u, w
.
u, v
=
u
T
v
for
u, v
∈
R
n
.
(i)
u, u
=
u
2
1
+
· · ·
+
u
2
n
≥
0. Thus
0
,
0
= 0
2
+
. . .
+ 0
2
= 0 and if
u, u
= 0,
then each
u
i
= 0, so
u
=
→
0 .
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 Spring '09
 RUDYAK
 Linear Algebra, ax, Hilbert space, inner product, Inner product space

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