# Finding the kernel of T

1. Dec 5, 2009

### hachi_roku

1. The problem statement, all variables and given/known data
let T: R^4 --->R^3, where T(v)=A(v) and matrix A is defined by

A = [2 1 -1 1
1 2 0 5
4 -1 1 0
Find kernel of T, nullity of T, range of T and rank of T

2. Relevant equations

3. The attempt at a solution
ok. ker(T) = Null(A)

[2 1 -1 1][v1]= [0]
[1 2 0 5] [v2]= [0]
[4 -1 1 0][v3] = [0]
[v4]

simplifying using gauss jordan:

1 0 0 1/6 [v1] [0]
0 1 0 29/12 [v2] = [0]
0 0 1 7/4 [v3] [0]
[v4]

v1 + 1/6V4 = 0
V2 + 29/12v4 = 0
v3 + 7/4v4 = 0 so [ -1/6v4]
[-29/12v4] V all real
[-7/4v4]
[v4]

range and rank i got 3

is this right?

2. Dec 6, 2009

anyone?

3. Dec 6, 2009

### HallsofIvy

Staff Emeritus
First, don't get upset if someone doesn't respond within four hours!

Second, it is easy to check for yourself if you have the kernel right: Take A of each of the vectors in the kernel and see if it is 0. Frankly, I don't know what by mean by
"[ -1/6v4]
[-29/12v4] V all real
[-7/4v4]
[v4]"
Do you mean that the kernel is the one dimensional subspace of R4 spanned by the single vector <-1/6, -12/12, -7/4, 1>?

Finally, no, the nullity and range cannot both be 3. By the "rank-nullity theorem", rank and nullity must add to the dimension of the domain space, in this case, 4.
If you are saying that the kernel is spanned by that single vector, then the kernel has dimension 1, so the nullity is 1, and the rank is 4-1= 3.

4. Dec 6, 2009

### hachi_roku

yes thats what i meant by writing that answer, so to check this, i take the matrix of that and see if it is zero?

5. Dec 7, 2009

### HallsofIvy

Staff Emeritus
Once again, "take this matrix and see if it is 0" makes no sense. Take what matrix and see if what "makes no sense"?