A First Course in Linear Algebra, 2017a

214 R n
Example 4.110: Null Space of A
Let
Find the null space of A.
A =
⎡
⎢
⎣
1 2 1 0 1
2 −1 1 3 0
3 1 2 3 1
4 −2 2 6 0
⎤
⎥
⎦
Solution. To find the null space, we need to solve the equation AX = 0. The augmented matrix and
corresponding reduced row-echelon form are given by
⎡
⎡
⎤
1 0 3 ⎤
6 1
5 5 5
0
1 2 1 0 1 0
⎢ 2 −1 1 3 0 0
⎥
⎣ 3 1 2 3 1 0 ⎦ →···→ 0 1 1 5
− 3 2 5 5
0
⎢
⎥
⎣
4 −2 2 6 0 0
0 0 0 0 0 0 ⎦
0 0 0 0 0 0
It follows that the first two columns are pivot columns, and the next three correspond to parameters.
Therefore, null(A) is given by
⎡ ( ) ( ) ( ) ⎤
−
3
5
s + −
6
5
t + −
1
( ) ( 5 r
−
1
5
s +
35
) ( )
t + −
2
5
r
⎢
s
⎥ : s,t,r ∈ R.
⎣
t
⎦
r
We write this in the form
⎡
s
⎢
⎣
− 3 5
− 1 5
1
0
0
⎤
⎡
+t
⎥ ⎢
⎦ ⎣
− 6 5
3
5
0
1
0
⎤
⎡
+ r
⎥ ⎢
⎦ ⎣
− 1 5
− 2 5
0
0
1
⎤
: s,t,r ∈ R.
⎥
⎦
In other words, the null space of this matrix equals the span of the three vectors above. Thus
⎧⎡
− 3 ⎤ ⎡
5
− 6 ⎤ ⎡
5
− 1 ⎤⎫
5
⎪⎨
− 1 3
5
5
− 2 5
⎪⎬
null(A)=span
⎢ 1
,
⎥ ⎢ 0
,
⎥ ⎢ 0
⎥
⎣ 0 ⎦ ⎣ 1 ⎦ ⎣ 0 ⎦
⎪⎩
⎪⎭
0 0 1
Notice also that the three vectors above are linearly independent and so the dimension of null(A) is 3.
The following is true in general, the number of parameters in the solution of AX = 0 equals the dimension
♠