# Showing that a basis for the ker(A) is in the kernel

1. Jan 21, 2008

### Jaqi Rose

1. The problem statement, all variables and given/known data
I'm new to this and I was wondering if any one could help me out

given:

x+z-w=1
y-z+w=1
x+y+z=3

find the coefficient matrix A, the vector of constants B, use Gauss-jordan elimination to solve the system. Find the Rank(A), the Null(A) and a basis for the im(A) and a basis for the ker(A) then verify that the vectors and in the kernel

2. Relevant equations

3. The attempt at a solution

I reduced the matrix, found the rank to be 3 and the null to be 1, the basis for the

im(A) =( [1;0;1] , [0;1;1] , [1;-1;1] )

then I get to the kernel, I set the equations from the reduced matrix to zero and found

x=w
y=-w
z=0

so the basis for the kernel is [1;-1;0;1] right?

how do I show that that basis is in the kernel?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 21, 2008

### Mathdope

If anything is in the kernel, then it maps to the zero vector. Does this happen?

Last edited: Jan 21, 2008
3. Jan 21, 2008

### Jaqi Rose

I'm not sure what " maps to the zero vector" means, this is my first course in Linear Algerbra ( two class periods so far ) and we don't have a book to work from.

After I reduce the matrix A|B I get

1 0 0 -1 | 0
0 1 0 1 | 2
0 0 1 0 | 1

which gives

x -w = 0
y+w = 2
z=1

i set these to zero ( not sure if that is right ) and then solve the equations for x and y
then i have

X = [x;y;z;w] = [1;-1;0;1]*w

that column vector then becomes my ker(A) right?

4. Jan 21, 2008

### Mathdope

"maps to the zero vector" means that if you plug the basis vector into your system you get zero.

So what I was getting at was - if you plug the basis vector that you got for the kernel in your system, does zero come out? If so, it's a basis for the kernel (assuming you did the rest correctly).

5. Jan 21, 2008

### Mystic998

That a vector v "maps to the zero vector" just means that A*v = 0. You can talk about A as a matrix, but you can also talk about it as function from one vector space to another. That's where the "mapping" terminology comes from.

6. Jan 21, 2008

### Jaqi Rose

yeah, if i take my basis for the ker(A) and plug the vaules back into the original system i get all of the equations to equal zero. is that right then?

thanks for the help by the way :)

7. Jan 21, 2008

### Mathdope

As long as the rest is correct - you have a kernel of dimension 1, you have a nonzero vector in it, so it is a basis. I haven't actually looked at the system itself.

8. Jan 21, 2008

Thanks!!!!!!