Download Math 311: Topics in Applied Math 1 3: Vector Spaces 3.2

Math 311: Topics in Applied Math 1
3: Vector Spaces
3.2: Subspaces
Solution
T
A is not a subspace of R3 since x = 1 0 0
∈A
T
but 2x = 2 0 0
∈
/ A because 2 + 0 6= 1.
B is a subspace of R3 , geometrically a line in 3-D space
T
through the origin. For let x = x x x
and
T
y= y y y
be in B and α ∈ R. Then
Summary
Below, V is a vector space over a field F of scalars.
T
• B is nonempty since 0 = 0 0 0
∈ B.
T
• αx = αx αx αx
∈ B.
T
• x+y = x+y x+y x+y
∈ B.
• A nonempty subset S ⊂ V closed under scalar
multiplication and vector addition is called a
subspace of V . I.e., S is a subspace provided
– 1. αx ∈ S whenever α ∈ F and x ∈ S,
C is a subspace of R3 , geometrically a plane in space that
T
contains the origin. For let x = x1 x2 x1 + x2
and
T
be in C and α ∈ R. Then
y = y1 y2 y1 + y2
– 2. x + y ∈ S whenever x, y ∈ S.
• Every subpsace S of V is a vector space in its own
right, having inherited the requisite stucture from V .
Accordingly, every subspace of V must contain the
zero vector of V .
T
• C is nonempty since 0 = 0 0 0
∈ C.
T
• αx = αx1 αx2 αx1 + αx2
∈ C.
• x+y =
• The null space N (A) of an m × n matrix A is the set
T
x1 + y1 x2 + y2 (x1 + y1 ) + (x2 + y2 )
∈ C.
of all solutions of the homogeneous system Ax = 0.
n
It forms a subspace of R , the vector space of all
T
n-element column vectors.
D is not a subspace of R3 since x = 1 0 1
∈D&
T
T
y= 0 1 1
∈ D but x + y = 1 1 2
∈
/D
n
because 2 6= 1.
• The set of all linear combinations ∑ αk vk where
k=1
αk ∈ F and vk ∈ V is called the span of v1 , . . . , vn ,
written Span (v1 , . . . , vn ). It is a subspace of V .
131/3
Determine whether the following are subspaces of R2×2 ,
the set of all real 2 × 2 matrices over the real field R.
• A set {v1 , . . . , , vn } is a spanning set of V if and
only if every vector in V can be written as a linear
combination of this spanning set.
(a) The set D of all 2 × 2 diagonal matrices
(c) The set L of all 2 × 2 lower triangular matrices
(e) The set B of all 2 × 2 matrices B such that b11 = 0
Examples
(g) The set S of all singular 2 × 2 matrices
131/2
Solution
Determine whether the following sets form subspaces of
R3 , the vector space of all 3-element real column vectors
over the real field R.
(To save vertical space, we’ll denote column vectors as
transposes of row vectors.)
n
o
T x1 x2 x3
A=
x1 + x3 = 1
2×2
(a) The
set D is asubspace of
R . For let
a11 0
b11 0
A=
and B =
0 a22
0 b22
be in D and α ∈ R. Then
1 0
• D is nonempty since
∈ D.
0 2
αa11
0
• αA =
∈ D.
0
αa22
a11 + b11
0
• A+B =
∈ D.
0
a22 + b22
o
T x
=
x
=
x
1
2
3
o
T x3
x3 = x1 + x2
B=
n
x1 x2 x3
C=
n
x1 x2
D=
n
x1 x2 x3
o
T x3 = x1 or x3 = x2
1
2×2
(c) The
let
set L is a subspace of R . For a11 0
b11 0
and B =
A=
a21 a22
b21 b22
be in L and α ∈ R. Then
1 0
• L is nonempty since
∈ L.
2 3
αa11
0
• αA =
∈ L.
αa21 αa22
a11 + b11
0
• A+B =
∈ L.
a21 + b21 a22 + b22
132/9a
1
0
. Determine the subspace S of R2×2
Let A =
0 −1
consisting of all 2 × 2 matrices that commute with A.
Solution
p q
∈ S. Then AB = BA, whence
Let B =
r s
p
q
p −q
=
. Thus q = −q and −r = r.
−r −s
r −s
In other words, 2q = 2r = 0, so q = r = 0. Accordingly,
S is the set of all 2 × 2 matrices having the form
p 0
, p, s ∈ R.
0 s
2×2
(e) The
set B is a subspace of R . For let
0 a12
0 c12
A=
and C =
a21 a22
c21 c22
be in B and α ∈ R. Then
0 1
• B is nonempty since
∈ B.
2 3
0
αa12
• αA =
∈ B.
αa21 αa22
0
a12 + c12
• A+C =
∈ B.
a21 + c21 a22 + c22
R2×2 .
1 2
0 0
In other words, the set of 2 × 2 diagonal matrices.
(g) S is not a subspace of
For A =
and
0 0
B=
are in S since their determinants are zero
3 4
1 2
and hence they are singular. But A + B =
∈
/S
3 4
since its detminant is −2 6= 0 and thus it is nonsingular.
131/6b
Recall that C [−1, 1] is the vector space of real continuous
functions on [−1, 1] over R. Determine whether the set S
of odd functions in C [−1, 1] is a subspace of C [−1, 1].
Solution
The set S is a subspace of C [−1, 1]. For let f , g ∈ S and
α ∈ R. Then
• S is nonempty since the function h defined by
h (x) = x on [−1, 1] is in S because
h (−x) = −x = −h (x) for x ∈ [−1, 1].
• α f ∈ S since for x ∈ [−1, 1], we have
(α f ) (−x) = α f (−x) = α (− f (x)) = − (α f (x)) = − (α f ) (x) ,
whence α f is an odd function.
• f + g ∈ S since for x ∈ [−1, 1], we have
( f + g) (−x) = f (−x)+g (−x) = − f (x)−g (x) = − ( f (x) + g (x)) = − ( f + g) (x) ,
whence f + g is an odd function.
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