Question about nullity and rank

  • Thread starter georgeh
  • Start date
  • #1
68
0
So i have the following matrix:
A= [2,0,-1; 4,0,-2;0,0,0]
I do r-r-e
I get
[1,0,-1/2;0,0,0;0,0,0]
So my rank for A is 1, because I only have 1 leading one.
Now for my nullity, i get the following
x_1 - 1/2 X_3 = 0
-->
x_1=1/2 x_3
therefore

[x_1,X_2,X_3] =[1/2;0;1]t
Which would imply that nullity = 1
which then wouldn't satisfy the Dimension theorem for matrices..
which states
rank(a)+nullity(a) = n, where n is the # of columns.
..
so my question is.. what am I doing wrong?
 

Answers and Replies

  • #2
113
1
How did you come up with x_2 = 0?
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
georgeh said:
So i have the following matrix:
A= [2,0,-1; 4,0,-2;0,0,0]
I do r-r-e
I get
[1,0,-1/2;0,0,0;0,0,0]
So my rank for A is 1, because I only have 1 leading one.
Now for my nullity, i get the following
x_1 - 1/2 X_3 = 0
-->
x_1=1/2 x_3
therefore

[x_1,X_2,X_3] =[1/2;0;1]t
Which would imply that nullity = 1
which then wouldn't satisfy the Dimension theorem for matrices..
which states
rank(a)+nullity(a) = n, where n is the # of columns.
..
so my question is.. what am I doing wrong?
The equation x1- (1/2)x3= 0 tells you that x3= 2x1 alright but tells you nothing about x2. A basis for the null space is [1, 0, 2] (equivalent to your [1/2, 0, 1]) and [0, 1, 0]. The nullity is 2.
 

Related Threads on Question about nullity and rank

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
5
Views
4K
  • Last Post
Replies
13
Views
3K
Replies
1
Views
3K
  • Last Post
Replies
9
Views
2K
Replies
7
Views
2K
Replies
4
Views
2K
Top