The solution sets of homogeneous linear systems provide an important source of vector spaces. Let
A be an
m by
n matrix, and consider the homogeneous system
Since
A is
m by
n, the set of all vectors
x which satisfy this equation forms a subset of
R
n
. (This subset is nonempty, since it clearly contains the zero vector:
x =
0 always satisfies
A
x =
0.) This subset actually forms a subspace of
R
n
, called the
nullspace of the matrix
A and denoted
N(A). To prove that
N(A) is a subspace of
R
n
, closure under both addition and scalar multiplication must be established. If
x1 and
x2 are in
N(A), then, by definition,
A
x1 =
0 and
A
x2 =
0. Adding these equations yields
which verifies closure under addition. Next, if
x is in
N(A), then
A
x =
0, so if
k is any scalar,
verifying closure under scalar multiplication. Thus, the solution set of a homogeneous linear system forms a vector space. Note carefully that if the system is
not homogeneous, then the set of solutions is
not a vector space since the set will not contain the zero vector.
Example 1: The plane
P in Example 7, given by 2
x +
y − 3
z = 0, was shown to be a subspace of
R3. Another proof that this defines a subspace of
R3 follows from the observation that 2
x +
y − 3
z = 0 is equivalent to the homogeneous system
where
A is the 1 x 3 matrix [2 1 −3].
P is the nullspace of
A.
Example 2: The set of solutions of the homogeneous system
forms a subspace of
R
n
for some
n. State the value of
n and explicitly determine this subspace.
Since the coefficient matrix is 2 by 4,
x must be a 4-vector. Thus,
n = 4: The nullspace of this matrix is a subspace of
R4. To determine this subspace, the equation is solved by first row-reducing the given matrix:
Therefore, the system is equivalent to
that is,
If you let
x3 and
x4 be free variables, the second equation directly above implies
Substituting this result into the other equation determines
x1:
Therefore, the set of solutions of the given homogeneous system can be written as
which is a subspace of
R4. This is the nullspace of the matrix
Example 3: Find the nullspace of the matrix
By definition, the nullspace of
A consists of all vectors
x such that
A
x =
0. Perform the following elementary row operations on
A,
to conclude that
A
x =
0 is equivalent to the simpler system
The second row implies that
x2 = 0, and back-substituting this into the first row implies that
x1 = 0 also. Since the only solution of
A
x =
0 is
x =
0, the nullspace of
A consists of the zero vector alone. This subspace, {
0}, is called the
trivial subspace (of
R2).
Example 4: Find the nullspace of the matrix
To solve
B
x =
0, begin by row-reducing
B:
The system
B
x =
0 is therefore equivalent to the simpler system
Since the bottom row of this coefficient matrix contains only zeros,
x2 can be taken as a free variable. The first row then gives
so any vector of the form
satisfies
B
x =
0. The collection of all such vectors is the nullspace of
B, a subspace of
R2:
Vector Algebra
Real Euclidean Vector Spaces
