8.3. THE RANK OF A MATRIX 157

The row reduced echelon form is 1 0 −9 9 20 1 5 −3 00 0 0 0 0

 .

Therefore, the rank of this matrix equals 2. All columns of this row reduced echelon formare in

span

 1

00

 ,

 010

 .

For example,  −950

=−9

 100

+5

 010

 .

By Lemma 8.2.5, all columns of the original matrix, are similarly contained in the span ofthe first two columns of that matrix. For example, consider the third column of the originalmatrix.  1

68

=−9

 113

+5

 237

 .

How did I know to use −9 and 5 for the coefficients? This is what Lemma 8.2.5 says!It says linear relationships are all preserved. Therefore, the column space of the originalmatrix equals the span of the first two columns. This is the desired efficient description ofthe column space.

What about an efficient description of the row space? When row operations are used,the resulting vectors remain in the row space. Thus the rows in the row reduced eche-lon form are in the row space of the original matrix. Furthermore, by reversing the rowoperations, each row of the original matrix can be obtained as a linear combination ofthe rows in the row reduced echelon form. It follows that the span of the nonzero rowsin the row reduced echelon matrix equals the span of the original rows. In the aboveexample, the row space equals the span of the two vectors,

(1 0 −9 9 2

)and(

0 1 5 −3 0).

Example 8.3.7 Find the rank of the following matrix and describe the column and rowspaces efficiently. 

1 2 1 3 21 3 6 0 21 2 1 3 21 3 2 4 0

 (8.2)