A First Course in Linear Algebra, 2017a

4.10. Spanning, Linear Independence and Basis in R n 213
Example 4.109: Null Space of A
Let
Find null(A) and im(A).
⎡
A = ⎣
1 2 1
0 −1 1
2 3 3
⎤
⎦
Solution. In order to find null(A), we simply need to solve the equation A⃗x =⃗0. This is the usual procedure
of writing the augmented matrix, finding the reduced row-echelon form and then the solution. The
augmented matrix and corresponding reduced row-echelon form are
⎡
⎣
1 2 1 0
0 −1 1 0
2 3 3 0
⎤
⎡
⎦ →···→⎣
1 0 3 0
0 1 −1 0
0 0 0 0
The third column is not a pivot column, and therefore the solution will contain a parameter. The solution
to the system A⃗x =⃗0 isgivenby ⎡ ⎤
−3t
⎣ t ⎦ : t ∈ R
t
which can be written as
⎡ ⎤
−3
t ⎣ 1 ⎦ : t ∈ R
1
Therefore, the null space of A is all multiples of this vector, which we can write as
⎧⎡
⎤⎫
⎨ −3 ⎬
null(A)=span ⎣ 1 ⎦
⎩ ⎭
1
Finally im(A) is just {A⃗x :⃗x ∈ R n } and hence consists of the span of all columns of A,thatisim(A)=
col(A).
Notice from the above calculation that that the first two columns of the reduced row-echelon form are
pivot columns. Thus the column space is the span of the first two columns in the original matrix,andwe
get
⎧⎡
⎨
im(A)=col(A)=span ⎣
⎩
Here is a larger example, but the method is entirely similar.
1
0
2
⎤
⎡
⎦, ⎣
2
−1
3
⎤
⎦
⎤⎫
⎬
⎦
⎭
♠