Let
A be a matrix. Recall that the dimension of its column space (and row space) is called the rank of
A. The dimension of its nullspace is called the
nullity of
A. The connection between these dimensions is illustrated in the following example.
Example 1: Find the nullspace of the matrix
The nullspace of
A is the solution set of the homogeneous equation
A
x =
0. To solve this equation, the following elementary row operations are performed to reduce
A to echelon form:
Therefore, the solution set of
A
x =
0 is the same as the solution set of
A′
x = 0:
With only three nonzero rows in the coefficient matrix, there are really only three constraints on the variables, leaving 5 − 3 = 2 of the variables free. Let
x4 and
x5 be the free variables. Then the third row of
A′ implies
The second row now yields
from which the first row gives
Therefore, the solutions of the equation
A
x =
0 are those vectors of the form
To clear this expression of fractions, let
t1 = ¼
x4 and
t2 = ½
x5 then, those vectors
x in
R5 that satisfy the homogeneous system
A
x =
0 have the form
Note in particular that the number of free variables—the number of parameters in the general solution—is the dimension of the nullspace (which is 2 in this case). Also, the rank of this matrix, which is the number of nonzero rows in its echelon form, is 3. The sum of the nullity and the rank, 2 + 3, is equal to the number of columns of the matrix.
The connection between the rank and nullity of a matrix, illustrated in the preceding example, actually holds for
any matrix:
The Rank Plus Nullity Theorem. Let
A be an
m by
n matrix, with rank
r and nullity ℓ. Then
r + ℓ =
n; that is,
Proof. Consider the matrix equation
A
x =
0 and assume that
A has been reduced to echelon form,
A′. First, note that the elementary row operations which reduce
A to
A′ do not change the row space or, consequently, the rank of
A. Second, it is clear that the number of components in
x is
n, the number of columns of
A and of
A′. Since
A′ has only
r nonzero rows (because its rank is
r),
n − r of the variables
x1,
x2, …,
xn
in
x are free. But the number of free variables—that is, the number of parameters in the general solution of
A
x = 0—is the nullity of
A. Thus, nullity
A =
n − r, and the statement of the theorem,
r + ℓ =
r + (
n −
r) =
n, follows immediately.
Example 2: If
A is a 5 x 6 matrix with rank 2, what is the dimension of the nullspace of
A?
Since the nullity is the difference between the number of columns of
A and the rank of
A, the nullity of this matrix is 6 − 2 = 4. Its nullspace is a 4-dimensional subspace of
R6.
Example 3: Find a basis for the nullspace of the matrix
Recall that for a given
m by
n matrix
A, the set of all solutions of the homogeneous system
A
x = 0 forms a subspace of
R
n
called the nullspace of
A. To solve
A
x = 0, the matrix
A is row reduced:
Clearly, the rank of
A is 2. Since
A has 4 columns, the rank plus nullity theorem implies that the nullity of
A is 4 − 2 = 2. Let
x3 and
x4 be the free variables. The second row of the reduced matrix gives
and the first row then yields
Therefore, the vectors
x in the nullspace of
A are precisely those of the form
which can be expressed as follows:
If
t1 = 1/7
x3 and
t2 = 1/7
x4, then
x =
t1(−2, −1, 7, 0)T +
t2(−4, 12, 0, 7)T, so
Since the two vectors in this collection are linearly independent (because neither is a multiple of the other), they form a basis for
N(A):
Vector Algebra
Real Euclidean Vector Spaces
