170 CHAPTER 8. RANK OF A MATRIX

Theorem 8.5.21 Let {v1, · · · ,vm} be a linearly independent set of vectors in Fn. Thenthere is a larger set of vectors, {v1, · · · ,vm,vm+1, · · · ,vn} which is a basis for Fn.

Example 8.5.22 The vectors,



1100

 ,

1010

 are linearly independent. Enlarge this

set of vectors to form a basis for R4.

Using the above technique, consider the following matrix.1 1 1 0 0 01 0 0 1 0 00 1 0 0 1 00 0 0 0 0 1

whose row reduced echelon form is

1 0 0 1 0 00 1 0 0 1 00 0 1 −1 −1 00 0 0 0 0 1

The pivot columns are numbers 1,2,3, and 6. Therefore, a basis is

1100

 ,

1010

 ,

1000

 ,

0001



8.5.5 Finding The Null Space Or Kernel Of A MatrixLet A be an m×n matrix.

Definition 8.5.23 ker(A), also referred to as the null space of A is defined as follows.

ker(A) = {x : Ax = 0}

and to find ker(A) one must solve the system of equations Ax = 0.

This is not new! There is just some new terminology being used. To repeat, ker(A) isthe solution to the system Ax = 0.

Example 8.5.24 Let

A =

 1 2 10 −1 12 3 3

 .

Find ker(A).