Welcome to a lesson
on a Null Space of a Matrix.
In this lesson,
we'll show how to check to see if a given vector
is in the null space of a matrix,
how to find the spanning set of a null space,
and also how to find a basis for a null space.
Let A be an m by n matrix.
Then the null space of A is the set of all vectors X in Rn
such that matrix A times vector x equals the 0 vector.
Which means to check to see
if a given vector is in the null space of a matrix,
we check the vector equation,
A times vector X equals 0,
where vector X would be the given vector.
If A times vector X equals a 0 vector,
then the vector is in the null space of the matrix.
If A times vector X is not equal to the 0 vector,
the vector is not in the null space of the matrix.
Let's take a look at an example.
Below we are given matrix A
and asked to determine if the following vectors
are in the null space of A.
Which means you need to check A times vector u
and A times vector v
to see if the products are equal to the 0 vector.
So for a vector u,
we have again matrix A times vector u,
and for vector v, we have A times vector v.
Next, we determine the products.
Notice A times vector u is equal to the 0 vector,
which indicates vector u is in the null space of matrix A.
Also notice how A times vector v
is not equal to the 0 vector,
which indicates vector v is not in the null space of A.
And now let's find the spanning set of a null space
and a basis for a null space.
To find the spanning set of a null space,
we solve the homogeneous vector equation,
A times vector X equals the 0 vector.
The spanning set contains all the vectors X in Rn
that satisfy the homogeneous vector equation.
The spanning set contains one vector for each free variable.
If there are no free variables when solving Ax equals 0,
then the only vector in the null space of A
is a 0 vector,
and Ax equals 0 has only the trivial solution.
Once we have the spanning set for the null space of A,
the vectors in the spanning set form a basis
for the null space of A.
Let's take a look at an example.
We are given matrix A
and asked to find the spanning set of the null space of A
and a basis for the null space of A.
We begin by setting up the equation matrix A times vector X
equals the 0 vector.
A couple of things to notice here,
matrix A is a 3 X 4 matrix.
When multiplying matrices,
the number of columns in the first matrix
must equal the number of rows in the second matrix.
Now, this vector x must have 4 components
written here as a 4 x 1 column matrix.
Also, notice how the product is equal to a 3 x 1 matrix
or, in our case, the 0 vector in R3.
So the vector is X on the left in R4,
and the 0 vector on the right is in R3.
The next step in solving the equation
is to write the corresponding augmented matrix.
The first row is -2, -3, 4, 5, 0.
The second row is 1, 2, 0, -1, 0.
And the third row is 2, 4, 0, -2, 0.
Next, we write the augmented matrix
in reduced row-echelon form,
which I've already done here on the right.
Let's go and label the first four columns,
X sub 1 through X sub 4.
We have pivots in row one column one and row two column two,
which indicates X sub 1 and X sub 2 are the basic variables,
and X sub 3 and X sub 4 are the free variables.
So because we have two free variables,
we know these spanning set will contain two vectors.
The first row indicates that X sub 1 minus 8 X sub 3
minus 7 X sub 4 is equal to 0
or X sub 1 is equal to 8 X sub 3
plus 7 X sub 4.
The second row indicates that X sub 2
plus 4 X sub 3 plus 3 X sub 4 equals 0
or X sub 2 is equal to -4 X sub 3
minus 3 X sub 4.
X sub 3 and X sub 4 are free variables.
X sub 3 is equal to X sub 3,
X sub 4 is equal to X sub 4.
So all the vectors that satisfy the equation,
matrix A times vector x equals the 0 vector
are the vectors X1, X2, X3, X4,
where X1 again is equal to 8 X sub 3
plus 7 X sub 4,
X sub two is equal to -4 x sub 3 minus 3 x sub 4,
and X sub 3 is equal to X sub 3,
X sub 4 is equal to X sub 4.
And now if this is linear combination of two vectors,
where we have X sub 3 times a vector formed
using the coefficients of X sub 3,
which are 8, - 4, 1, 0
plus X sub 4 times a vector formed
using the coefficients of X sub 4,
which are 7, -3, 0, 1.
This is the information we need
in order to find the spanning set of the null space of A,
as well as a basis for the null space of A.
Since the solutions to the equation matrix A
times vector X equals the 0 vector,
is equal to the linear combinations
of the vectors 8, -4, 1, 0,
and 7, -3, 0, 1,
we can say the null space
equals the span of these two vectors,
and relatively, we can also say,
a basis for the null space of A
is the set of these two vectors.
These two statements are related,
but by saying that the null space of A
equals the span of these two vectors,
all the linear combinations of these two vectors
make up the null space of A,
and by stating a basis for the null space of A
is equal to the set of these two vectors,
these two vectors span the null space and are independent,
and any vector in the null space of A can be written
as a linear combination of these two vectors.
I hope you found this helpful.