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 *x* _{4} and *x* _{5} be the free variables. Then the third row of *A*′ implies

The second row now yields

Therefore, the solutions of the equation *A* **x** = **0** are those vectors of the form

To clear this expression of fractions, let *t* _{1} = ¼ *x* _{4} and *t* _{2} = ½ *x* _{5} then, those vectors **x** in **R** ^{5} 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,

rank *A* + nullity *A* = the number of columns of *A*

*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 *x* _{1}, *x* _{2}, …, *x _{n} *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 **R** ^{6}.

**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 *x* _{3} and *x* _{4} be the free variables. The second row of the reduced matrix gives

Therefore, the vectors **x** in the nullspace of *A* are precisely those of the form

If *t* _{1} = 1/7 *x* _{3} and *t* _{2} = 1/7 *x* _{4}, then **x** = *t* _{1}(−2, −1, 7, 0) _{T} + *t* _{2}(−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)*: