ECE 275A Homework # 2 Solutions
1. ECE 174 Midterm solutions are in a separate file.
2. Let
V
and
W
be independent (disjoint) subspaces of a vector space
X
, such that
X
=
V
+
W
.
1
Consider any two vectors
x
=
x
1
+
x
2
and
y
=
y
1
+
y
2
in
X
, where
x
1
, y
1
∈ V
and
x
2
, y
2
∈ W
give the unique decomposition of
x
and
y
along the
companion subspaces
V
and
W
respectively.
Let
P
=
P
VW
denote the projection
operator which projects
X
onto
V
along
W
. Then
P x
=
x
1
and
P y
=
y
1
. For any
two scalars
α
and
β
we have
αx
+
βy
=
α
(
x
1
+
x
2
) +
β
(
y
1
+
y
2
) = (
αx
1
+
βy
1
) + (
αx
2
+
βy
2
)
∈ X
,
with (
αx
1
+
βy
1
)
∈ V
and (
αx
2
+
βy
2
)
∈ W
since
V
and
W
are vector subspaces.
Therefore
P
(
αx
+
βy
) =
αx
1
+
βy
1
=
αPx
+
βPy ,
showing that a (possibly nonorthogonal) projection operator is linear.
3. Recall the the columns of [
V W
] are a basis for
X
iff they form a spanning set of vectors
for
X
which are also linearly independent. Also recall that a linear operator
P
is a
projection operator iff it is idempotent,
P
=
P
2
.
(a) The complementary subspace condition
X
=
V ⊕W
and the fact that the columns
of
V
and and the columns of
W
each respectively forms a basis for
V
and
W
implies
that for each
x
∈ X
there exists a
v
=
V α
∈ V
and a
w
=
Wβ
∈ W
such that
x
=
v
+
w
=
V α
+
Wβ
= [
V W
]
(
α
β
)
showing that the columns of [
V W
] span
X
.
The complementary subspace condition
X
=
V ⊕ W
implies that
V
and
W
are
disjoint, which is true iff
V ∩ W
=
{
0
}
.
Now suppose that the columns of
V
and
W
taken together
are
not
linearly independent. Then there exists
α
̸
= 0 and
β
̸
= 0 such that 0 =
V α
+
Wβ
.
(Recall that the columns of
V
are a linearly
independent set, as are the columns of
W
.) This yields
V α
=
−
Wβ
̸
= 0. But
V α
∈ V
and
−
Wβ
∈ W
, implying that
0
̸
=
V α
∈ V ∩ W
which contradicts the assumption that
V
and
W
are disjoint,
V ∩ W
=
{
0
}
.
Therefore the assumption that the columns of
V
and
W
taken together are not
linearly independent must be incorrect.
1
I.e., let
V
and
W
be
companion
subspaces of
X
.
1
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(b) Note that to show that a linear operator is a projection operator onto
V
we need
to show both that it is
idempotent
(i.e., that it is indeed a
projector
) and that
its
range is
V
(i.e., that it is a projector
onto
V
). Let
n
be the dimension of
X
. With
the columns
M
= [
V W
] linearly independent,
M
is an
n
×
n
invertible matrix.
Consider the matrix
P
=
M
(
I
0
0
0
)
M

1
= [
V
0]
M

1
.
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 Fall '08
 Chandrasekara
 Linear Algebra, Vector Space, inner product, A∗

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