Matrix A reduces to this first matrix,
which is in row echelon form.
I've also written the same matrix
in reduced row echelon form here on the far right.
We're asked to find the rank of A
as well as the dimension of the null space of A.
The dimension of a subspace is the number of basis vectors.
The rank of A is equal to the dimension
of the column space of A,
which is equal to the number of vectors
in a basis of the column space of A.
This is equal to the number of pivot columns
or the number of basic variables.
The nullity of A, or the dimension of the null space of A,
equals the number of vectors in a basis
of the null space of A.
Which is equal to the number of free variables
or the number of non pivot columns.
This leads us to the rank theorem which states,
the rank of A plus the dimension of the null space of A,
is equal to n where n is number of columns in the matrix.
And this should make sense because the rank of A
is equal to the number of pivot columns,
and the dimension of the null space of A
is equal to the number of non pivot columns.
If we sum the number of pivot columns and non pivot columns,
it'll give us a number of columns in matrix A.
So it doesn't matter whether we use row echelon form
or reduced row echelon form to identify
the pivot columns and non pivot columns.
Let's begin by identifying the pivot columns.
We have pivots in row one column one
and row two column three in both forms.
This indicates that column one
and column three are pivot columns.
Because we have two pivot columns,
the rank of matrix A is equal to two,
which indicates there are two vectors
in a basis of the column space of A.
And this should make sense because remember,
to find a basis for the column space of A,
the pivot columns or independent columns of matrix A
form a basis for the column space of A.
And we find the pivot columns or independent columns
by solving the equation A times vector X equals zero.
To do this, you write matrix A in row echelon form
or reduced row echelon form.
If we wanted to find a basis for the column space though,
we would have to go back to the original matrix A
and identify column one and column three
as the two independent vectors that span the column space.
And now let's find dimension of the null space of A.
To do this, we determine the number of non pivot columns.
We have column one and column three are the pivot columns.
Column two and column four are the non pivot columns.
Notice we don't have pivots in these columns.
This indicates the dimension of the null space
is equal to two, because we have two non pivot columns,
or because we have two free variables.
Notice two plus two is equal to four,
which is the number of columns of matrix A.
And again, if you wanted to find the vectors
that form a basis for the null space,
remember, we would solve the vector equation
A times vector X equals zero in the spanning set
of the solution set form a basis for the null space.
Which is why the number of free variables
or number of non pivot columns indicate
how many vectors are in a basis.
Let's take a look at one more example.
Here we're given matrix A in row echelon form,
and I also found reduced row echelon form.
Either form works for identifying the rank of A
and the dimension of the null space of A.
To determine the rank, we identify the pivot columns.
We have pivots in row one column one,
row two column two and row three column three in both forms.
We have three pivot columns column one,
column two and column three.
Which means the rank of A is equal to three.
And again this indicates there are three vectors
in a basis of the column space of A.
And if there are three pivot columns
and there are four total columns,
there's one non pivot column, column four.
Which indicates there is one free variable,
and the dimension of the null space of A is equal to one.
There is one vector in a basis of the null space of A.
I hope you found this helpful.