Subspaces of Rn
Consider the collection of vectors
The endpoints of all such vectors lie on the line
y = 3
x in the
x-y plane. Now, choose any two vectors from
V, say,
u = (1, 3) and
v = (-2, -6). Note that the sum of
u and
v,
is also a vector in
V, because its second component is three times the first. In fact, it can be easily shown that the sum of
any two vectors in
V will produce a vector that again lies in
V. The set
V is therefore said to be
closed under addition. Next, consider a scalar multiple of
u, say,
It, too, is in
V. In fact,
every scalar multiple of any vector in
V is itself an element of
V. The set
V is therefore said to be
closed under scalar multiplication.
Thus, the elements in
V enjoy the following two properties:
- Closure under addition:
The sum of any two elements in V is an element of V.
- Closure under scalar multiplication:
Every scalar multiple of an element in V is an element of V.
Any subset of
R n that satisfies these two properties—with the usual operations of addition and scalar multiplication—is called a
subspace of R n or a
Euclidean vector space. The set
V = {(
x, 3
x):
x ∈
R} is a Euclidean vector space, a subspace of
R2.
Example 1: Is the following set a subspace of
R2?
To establish that
A is a subspace of
R2, it must be shown that
A is closed under addition and scalar multiplication. If a counterexample to even one of these properties can be found, then the set is not a subspace. In the present case, it is very easy to find such a counterexample. For instance, both
u = (1, 4) and
v = (2, 7) are in
A, but their sum,
u +
v = (3, 11), is not. In order for a vector
v = (
v1,
v2 to be in
A, the second component (
v2) must be 1 more than three times the first component (
v1). Since 11 ≠ 3(3) + 1, (3, 11) ∉
A. Therefore, the set
A is not closed under addition, so
A cannot be a subspace. [You could also show that this particular set is not a subspace of
R2 by exhibiting a counterexample to closure under scalar multiplication. For example, although
u = (1, 4) is in
A, the scalar multiple 2
u = (2, 8) is not.]
Example 2: Is the following set a subspace of
R3?
In order for a sub
set of
R3 to be a sub
space of
R3, both closure properties (1) and (2) must be satisfied. However, note that while
u = (1, 1, 1) and
v = (2, 4, 8) are both in
B, their sum, (3, 5, 9), clearly is not. Since
B is not closed under addition,
B is not a subspace of
R3.
Example 3: Is the following set a subspace of
R4?
For a 4-vector to be in
C, exactly two conditions must be satisfied: Namely, its second component must be zero, and its fourth component must be −5 times the first. Choosing particular vectors in
C and checking closure under addition and scalar multiplication would lead you to conjecture that
C is indeed a subspace. However, no matter how many specific examples you provide showing that the closure properties are satisfied, the fact that
C is a subspace is established only when a general proof is given. So let
u = (
u1, 0,
u3, −5
u1) and
v = (
v1, 0,
v3, −5
v1) be arbitrary vectors in
C. Then their sum,
satisfies the conditions for membership in
C, verifying closure under addition. Finally, if
k is a scalar, then
is in
C, establishing closure under scalar multiplication. This proves that
C is a subspace of
R4.
Example 4: Show that if
V is a subspace of
R n , then
V must contain the zero vector.
First, choose any vector
v in
V. Since
V is a subspace, it must be closed under scalar multiplication. By selecting 0 as the scalar, the vector 0
v, which equals
0, must be in
V. [Another method proceeds like this: If
v is in
V, then the scalar multiple (−1)
v = −
v must also be in
V. But then the sum of these two vectors,
v + (−
v) =
0, mnust be in
V, since
V is closed under addition.]
This result can provide a quick way to conclude that a particular set is not a Euclidean space.
If the set does not contain the zero vector, then it cannot be a subspace. For example, the set
A in Example 1 above could not be a subspace of
R2 because it does not contain the vector
0 = (0, 0). It is important to realize that containing the zero vector is a
necessary condition for a set to be a Euclidean space, not a
sufficient one. That is, just because a set contains the zero vector does not guarantee that it is a Euclidean space (for example, consider the set
B in Example 2); the guarantee is that if the set does
not contain
0, then it is
not a Euclidean vector space.
As always, the distinction between vectors and points can be blurred, and sets consisting of points in
R n can be considered for classification as subspaces.
Example 5: Is the following set a subspace of
R2?
As illustrated in Figure
1 , this set consists of all points in the first quadrant, including the points (
x, 0) on the
x axis with
x > 0 and the points (0,
y) on the
y axis with
y > 0:
The set D is closed under addition since the sum of nonnegative numbers is nonnegative. That is, if ( x1, y1) and ( x2, y2) are in D, then x1, x2, y1, and y2 are all greater than or equal to 0, so both sums x1 + x2 and y1 + y2 are greater than or equal to 0. This implies that
However,
D is not closed under scalar multiplication. If
x and
y are both positive, then (
x, y) is in
D, but for any negative scalar
k,
since
kx < 0 (and
ky < 0). Therefore,
D is not a subspace of
R2.
Example 6: Is the following set a subspace of
R2?
As illustrated in Figure 2 , this set consists of all points in the first and third quadrants, including the axes:
The set E is closed under scalar multiplication, since if k is any scalar, then k(x, y) = ( kx, ky) is in E. The proof of this last statement follows immediately from the condition for membership in E. A point is in E if the product of its two coordinates is nonnegative. Since k2 > 0 for any real k,
However, although E is closed under scalar multiplication, it is not closed under addition. For example, although u = (4, 1) and v = (−2, −6) are both in E, their sum, (2, −5), is not. Thus, E is not a subspace of R2.
Example 7: Does the plane P given by the equation 2 x + y − 3 z = 0 form a subspace of R3?
One way to characterize P is to solve the given equation for y,
and write
If
p1 = (
x1, 3
z1 − 2
x1,
z1) and
p2 = (
x2, 3
z2 − 2
x2,
z2) are points in
P, then their sum,
is also in
P, so
P is closed under addition. Furthermore, if
p = (
x, 3
z − 2
x, z) is a point in
P, then any scalar multiple,
is also in
P, so
P is also closed under scalar multiplication. Therefore,
P does indeed form a subspace of
R3. Note that
P contains the origin. By contrast, the plane 2
x +
y − 3
z = 1, although parallel to
P, is
not a subspace of
R3 because it does not contain (0, 0, 0); recall Example 4 above. In fact, a plane in
R3 is a subspace of
R3 if and only if it contains the origin.
