# Finding null space of a given matrix

1. Oct 27, 2008

### Maxwhale

1. The problem statement, all variables and given/known data

Find null space of A, NS(A) and sketch NS(A) in R2 or R3.

A = [1 3 2; 2 6 4]
2. Relevant equations

AX = 0

3. The attempt at a solution

I know the second row is twice the first one. I tried to solve for x1, x2 and x3 putting everything in the form of AX =0. I did not get a confident answer to sketch.

2. Oct 27, 2008

### morphism

What did you get?

3. Oct 27, 2008

### Maxwhale

since there are three variables, x1, x2 and x3, i reduced them to row echelon presuming that x3=0 and then got x1=0 and x3=0. Is that a right way?

4. Oct 27, 2008

### morphism

What do you mean? The null space of A is a set.

5. Oct 27, 2008

### HallsofIvy

Staff Emeritus
Either say NS(A) is the set of vectors (x, y, z) such that .... or, perhaps simpler, give a basis for the Null Space.

6. Oct 27, 2008

### FourierX

We are not that far yet. We have not covered basis. I would really appreciate if you could put it simply so that i could understand. Please :)

7. Oct 27, 2008

### HallsofIvy

Staff Emeritus
You cannot assume x3 equals any specific number. The nullspace of A is all vectors <x1, x2, x3> such that
$$\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 6 & 4 \end{array}\right]\left[\begin{array}{c} x1 \\ x2 \\ x3\end{array}\right]= \left[\begin{array}{c}0 \\ 0 \\ 0 \end{array}\right]$$

You can row reduce that as you said you did: the bottom row will be 0, of course, the top row [1 3 2 0] which corresponds to the equation x1+ 3X2+ 2X3= 0. Since that is single equation in 3 unknown numbers, you can choose two of them to be whatever you want and solve for the third. If x2= 1 and x3= 0, what is x1?. If x2= 0 and x3= 1, what is x1?

What is the dimension of NP(A)? What is a basis for NP(A).
You can,

8. Oct 27, 2008

### only_huce

The matrix A which you gave has one pivot column making the other two columns free.

n-r = 3-1 = 2, the dimension of the nullspace is 2.

Set the free variables to 1 and 0, and use back substitution to find the missing variable in your nullspace vector. You should get:

s_1 = [-3, 1, 0] and s_2 = [-2, 0, 1], with dimension 2.

I'm pretty sure that's right.

9. Oct 28, 2008

### HallsofIvy

Staff Emeritus
If you do not know what a basis is then I guess that by "find the null space" you are meant to just write the equation satisfied by points in the null space. As you said, the two equations are equivalent so any such point must satisfy x+ 3y+ 2z= 0.

What is the graph of that?