Finding null space of a given matrix

[Maxwhale]

Homework Statement

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

A = [1 3 2; 2 6 4]

Homework Equations

AX = 0

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.

 

[morphism]

What did you get?

 

[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?

 

[morphism]

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

 

[HallsofIvy]

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

 

[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 :)

 

[HallsofIvy]

Homework Statement

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

A = [1 3 2; 2 6 4]

Homework Equations

AX = 0

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.

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?

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,

 

[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.

 

[HallsofIvy]

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 :)

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?