# Kernel/Range basis

1. Apr 30, 2010

### newtomath

L: R^4 => R^3 is defined by L(x,y,z,w) = (x+y, z+w, x+z)

A) Find a basis for ker L

We can re write L(x,y,z,w) as x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0).
I then reduced it to row echelon form

We now have the equations X-W=0 , Y+W=0, Z+W=0.

There are infinitely many solutions as X=W, Y= -W and Z=-W. So if we set W=1 we have

the basis for the kernel=Vector(1,-1, -1,1)

B) find a basis for range L

Given
L(x,y,z,w) = (x+y, z+w, x+z)

We can re write L(x,y,z,w) as x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0).
S= {(1,01) ,(1,0,0) ,(0,1,1),(0,1,0)} It spans L.

To find the basis for L we set {x* (1,01) + y *(1,0,0) + z*(0,1,1) + w*(0,1,0)} = 0,0,0

I reduced it and the leading one's appear in the first 3 columns of the reduced form, the first 3 vectors in the original matrix became a basis for the range of L
They are:
Vector( {(1, (0, (1})
,
Vector( 1, 0, 0})
,
Vector( 0, 0, 1})

C) verify theorem 10.7 (dim(KerL) + dim(rangeL) = dim V

The dim can be viewed as the # of vectors in of the Ker/range.

Given (dim(KerL) + dim(rangeL) = dim V we have 1+3=4, which is the number of dimensions in the original space (L(x,y,z,w)).

2. Apr 30, 2010

### Staff: Mentor

3. Apr 30, 2010

### newtomath

I forgot to type it. I believe A and B to be correct, but is my explanantion in C suffice?

4. Apr 30, 2010

### Staff: Mentor

Sure, it's fine, and the other parts are fine also.