Welcome to a lesson
on the column space of a matrix.
Let A be an M by N matrix with columns equal to the vectors
A sub one through A sub N,
the column space of A is equal to the span of the vectors
A sub one through A sub N.
The column space of A is a subspace of RM
where notice M is the number of rows in matrix A,
the column space of A contains all the linear combinations
of the vectors A sub one through A sub N,
which means the column space of A is in the form shown here
where C sub one through C sub N are scalars.
As an example,
let's consider the given three by four matrix.
In this matrix, vector A sub one is the first column,
vector A sub two is the second column,
vector A sub three is the third column,
and vector A sub 4 is the fourth column.
Notice the vectors have three components
and therefore the vectors are in R3,
the matrix has three rows.
The column space of this matrix is equal to the span
of these four vectors.
The column space of A, is a set of all vectors
of the form A times vector X, since A times vector X
is a linear combination of the columns of matrix A.
Sometimes a column space of A will span all of RM,
this occurs when A times vector X equals vector V,
has a solution for all of the vectors V in RM.
This occurs when A has a pivot in each row.
There are two common types of questions,
when working with the column space of a matrix,
number one, we will want to check to see
if the given vector V, is in the calm space of a matrix,
to do this, we solve the equation A times vector X
equals vector V, if there's at least one solution,
then the vector V is in the column space,
if there is no solution
the vector is not in the column space.
We will also want to find a basis for a column space,
the pivot columns of the matrix A
form a basis for the column space of A,
we find the pivot columns by solving the equation
A times vector X equals a zero vector,
the basis will be the corresponding column vectors
from the original matrix A,
not from the columns after performing row operations,
let's take a look at some examples.
Here we're given matrix A and asked to determine,
if the following vectors are in the column space of A,
to determine whether vector U is in the column space
of matrix A, we need to see if there is a solution
to the equation A times vector X equals vector U.
So we have the vector equation shown here,
followed by the augmented matrix,
and then the augmented matrix written in
reduce row echelon form.
Let's go and label the first three columns.
Notice in row three, we have zero, zero, zero, one,
we have a contradiction,
this row indicates that zero equals one,
which is never true, which means there's no solution
to the equation and therefore vector U
is not in the column space of A.
To check to see whether vector V
is in the column space of A,
we need to see if we have a solution to A times vector X
equals vector V, here we have the matrix equation,
the augmented matrix and then the augmented matrix
in reduced row echelon form.
Again let's label the columns, X1, X2, X3,
notice in row three we have a row of zeros
which indicates we have an infinite number of solutions
or a dependent system so we can actually go ahead
and just stop here because we have one or more solutions,
we know there is a solution to the equation
A times vector X equals vector V,
that's all we need to know vector V
is in the column space of A.
And now let's find a basis for a column space.
We're given matrix A and asked to find a basis
for the column space of A.
So again the pivot columns of matrix A,
form a basis for the column space of A,
to find the pivot columns,
we need to solve the homogeneous vector equation,
A times vector X equals the zero vector.
The basis will be the corresponding column vectors
from the original matrix A,
so we set up the homogeneous vector equation,
I've already written the augmented matrix,
and then written the augmented matrix
in reduced row echelon form,
the next step is to find the pivot columns,
we can see we have a pivot in row one column one,
as well as row two column two,
which means the pivot columns are column one and column two,
so if we go back to matrix A, the vectors in column one
and column two make up a basis for the column space of A.
A basis for the column space of A,
is the vector two, three, negative four
and one, zero, three.
We can also state that the column space of A,
is equal to the span of these two vectors.
I hope you found this helpful.