Welcome to a lesson
on how to determine the fundamental subspaces of a matrix.
Let A be an m by n matrix.
The fundamental subspaces of A are:
The null space of A,
which is equal to the set of all vectors x
such that matrix A times vector x equals a zero vector.
The null space of A transpose,
which is equal to the set of all vectors x
such that A transpose times vector x equals a zero vector.
The column space of A,
which is equal to the span of the columns of A,
and the column space of A transpose,
which is equal to the span of the columns of A transpose.
This is also equal to the span of the rows of A
or the row space of A.
For review, to find a basis for the null space of A,
we solve matrix A times vector x equals a zero vector.
The vectors in the spanning set
form a basis form the null space.
The number of free variables indicate how many vectors
are in a basis.
To find a basis for the column space of A,
the pivot columns of matrix A form basis
for the column space of A.
We find the pivot columns by solving matrix A times vector x
equals a zero vector.
A basis will be the corresponding column vectors
from the original matrix A.
Let's take a look at an example.
Let's determine the fundamental subspaces
of the given matrix A, which is a three by four matrix.
Let's begin by determining the null space of A
as well as the column space of A.
To do this, we begin by setting up the equation matrix A
times vector x equals a zero vector.
Notice here I've written the augmented matrix
for the equation
where the augmented matrix is a three by five matrix.
The fifth column is a column of zeros
from the zero vector on the right side of the equation.
Writing this augmented matrix in reduced row echelon form,
notice how we have pivots in column one and column in three.
So if we go back to matrix A,
column one and column three
form our basis for the column space.
So the vector one, zero, zero,
and the vector three, three, four
form our basis for the column space of matrix A.
And therefore, we can say the column space of matrix A
is equal to the span of these two vectors.
Now let's continue solving.
The first row of the augmented matrix
in reduced row echelon form indicates x one plus two x two
minus 1/2 x four equal zero.
The second row indicates x three
plus 1/2 x four equals zero.
Notice how there are no pivots in column two and column four
indicating x two and x four are free variables.
Let's parametrize the solution by letting x two equal s
and x four equal t.
Solving the first equation for x one,
we have x one equals negative three s plus 1/2 t.
Solving the second equation for x three,
we have x three equals negative 1/2 t.
And again, we still have x two equals s and x four equals t.
And therefore these solutions to the equation,
matrix A times vector x equals a zero vector
are all the vectors x in the form of s
times the vector negative three, one, zero, zero.
Notice how these entries come from the coefficients of s
in order from x one to x four,
and then we have plus t times the vector 1/2,
zero, negative 1/2, one,
where these entries are the coefficients of t
in order from x one to x four.
Which means the basis for the null space of matrix A
would be the vectors negative three, one, zero, zero,
and 1/2, zero, negative 1/2, one,
or we can say the null space of matrix A
is equal to the span of these two vectors.
And next we're going to determining the column space
of A transpose as well as the null space of A transpose.
Let's first find A transpose.
Remember to find A transpose, we take the rows of matrix A
and make them the columns of A transpose.
And now the next step is to set up the equation,
A transpose times vector x equals a zero vector.
Write the augmented matrix,
and write the augmented matrix in reduced row echelon form,
which I've already set up here on the right.
Notice how the augmented matrix is a four by four matrix,
where the fourth column is a column of zeros,
because again, of the zero vector
on the right side of the equation.
And now looking at the augmented matrix
in reduced row echelon form,
notice how we have pivots in column one and column two,
which means column one and column two of A transpose
are a basis for the column space of a transpose,
which would be the vectors one, three, three, one,
and zero, zero, four, two.
And therefore we can say the column space of A transpose
or the row space of A
is equal to the span of these two vectors.
And now we continue to solve the first row
of the augmented matrix in reduced row echelon form
indicates x one equals zero.
The second row indicates x two equals zero.
There is no pivot in column three.
x three is a free variable.
And therefore we write x three equals x three.
Parameterizing the solution, let the x three equal t,
and therefore all these solutions to the equation,
A transposed times vector x equals a zero vector
are all the vectors x
in the form of t times the vector zero, zero, one.
Which means the vector zero, zero, one
forms a basis for the null space of A transpose,
or we can say the null space of a transpose
is equal to the span of the vector zero, zero, one.
I hope you found this helpful.