As an example, let's use the matrix from the previous subsection:  From this row reduction we can see that the column space (image) has the following basis:  The pivotal columns of matrix are the pivotal rows of so the rows with pivots in the original matrix form the basis of the row space (coimage):  For the null space (kernel), we solve the system . This is, of course, equivalent to simply row reducing since the constant vector is . First convert the matrix above back into equation form:   Next, we set and and parameterize the solution as follows:     Thus, we can write the basis for the null space (kernel) as the following:  We will not show the basis for the left nullspace as it is rarely used in practice, but it can be found by applying the same method for finding the null space to .