Every matrix A defines 3 vector spaces -- the row space, the column space, and the nullspace. This lesson covers how to find a basis for the nullspace of a matrix or system of equations.

Theorem: The vectors that define the solution set of Ax = 0 form a basis for the nullspace of A.
The dimension (number of vectors) of the nullspace is called the nullity of A.

In order to find a basis for the nullspace, we solve the homogenous system and write the solution set in vector form.

Example 1: Find a basis for the nullspace of this homogenous system of equations:

Solution

The augmented matrix for this system is

, which we reduce to row-echelon form to get

, so we know that

It is obvious that x ₄ = 0 and since we have 3 equations in 5 unknowns, we need 2 parmeters. If we set x ₅ = t, then x ₃ = – t. Similarly, we set x ₂ = s, which makes x ₁ = – s – t.

The general solution for this system is: where s, t R.

This can be expressed as the sum of these 2 vectors:

where s, t R.

These 2 vectors, therefore, form a basis for the nullspace of A.
It is a 2-dimensional subspace of R⁵ since there are 2 vectors with 5 components.
This tells us that the rank of A is 3. There are 3 vectors in a basis for the row and column spaces of this matrix.

If the system has only the trivial solution, the zero vector forms a basis for the nullspace and the nullity is = 0.

.

Practice

1) Find a basis for and state the dimension of the nullspace of these homogeneous systems.

a)	b)
c)	d)

Solutions:

1) a) To set up the augmented matrix:

first, switch row2 with row1, and switch row3 with row2 to get:

, now (row3 – 2row1)/3 and ½(row3 + row2) to get:

, so the system has only the trivial solution x₁ = x₂ = x₃ = 0.

A basis for the nullspace of A is the 3-dimensional zero vector , nullity = 0

** Since A is a 3 × 3 square matrix, if , we know the system has only the trivial solution. Matrix A is row equivalent to I _{3 × 3}

b) The augmented matrix .

Since we have 4 variables and 2 equations, we need 2 parameters:

Let x₄ = t and let x₃ = 4s, which means that x₂ = – t – s and x₁ = – s.

This can be expressed as the sum of these 2 vectors:

where s, t R.

It is a 2-dimensional subspace of R⁴ since there are 2 vectors with 4 components.

c) To create the augmented matrix, switch the 1st and 2nd equations,
then use ½ (eq. 1) for row2 to get:

, now use row1 to eliminate entries in the first column.

step 2: (row3 – 2row1) and (row4 + 2row1)

step 3: (row3 – 3row2) and (row4 – row2)

Now we've got , so we add 10 × row3 to row 4 to get zeros in the last row.

When we set y = s, we get

w = s

x = – s

y = s

z = 0

s R

This can be expressed as the vector: so nullity = 1.

d) To set up the augmented matrix:

first, switch row3 with row1 to get:

, now (row2 + row1) and (row3 – 2row1) to get:

, now add row3 to row 2 to get ,

and when we multiply row2 by 1/3 and row3 by – 1/10, we get the reduced matrix

, which means the system has only the trivial solution x₁ = x₂ = x₃ = 0.

A basis for the nullspace of A is the 3-dimensional zero vector , nullity = 0

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