The Nullspace of a Matrix
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 x 1 and x 2 are in N(A), then, by definition, A x 1 = 0 and A x 2 = 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 R 3. Another proof that this defines a subspace of R 3 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 Rn 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 x 3 and x 4 be free variables, the second equation directly above implies
Substituting this result into the other equation determines x 1:
Therefore, the set of solutions of the given homogeneous system can be written as
which is a subspace of R 4. 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 x 2 = 0, and back‐substituting this into the first row implies that x 1 = 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 R 2).
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, x 2 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 R 2:
