What is the Kernel, Nullity, Range, and Rank of T given a specific matrix A?

[hachi_roku]

Homework Statement

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

Homework Equations

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?

 

[hachi_roku]

anyone?

 

[HallsofIvy]

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.

 

[hachi_roku]

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

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.

 

[HallsofIvy]

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