# Linear Algebra - vector spaces

1. Oct 29, 2009

### underacheiver

1. The problem statement, all variables and given/known data
part a.
Use the matrix A =
{[1,-1,0]
[0,-1,1]
[-1,2,-1]}
to compute T(x) for x =
{[1]
[2]
[3]}
Here, T:R^3-->R^3 is defined as T(x)=Ax.

part b.
describe the kernel of the transformation.

part c.
what is the nullity of the tarnsformation

part d.
what is the rank of the transformation

2. Relevant equations
I guess the only equation is the rank nullity theorum, and T(x)=Ax

3. The attempt at a solution
a.
i multiplied the matrix A and x and got
T(x)=
{[-1]
[1]
[0]}
this part wasn't too hard.

b.
the kernel is the solution of Ax=0, right?
i solved the matrix equation and got ker T=
{[t]
[t]
[t]}
Is this correct?

c. the nullity is the dimension of the kernel space, so is it 1?? (i am guessing)

d. the dimension of the space is 3, so the rank is 2?? according to the rank nullity theorum?? (once again, guessing)

I'm sure i am doing something wrong. what is the logic behind parts c and d?

2. Oct 29, 2009

### lanedance

looks reasonable to me, this represents all vectors on the line t.(1,1,1) for any number t
yes so you found the kernal space is line, which has dimension 1
so the rank is 2, for a linear transformation, the rank is equal to the dimension of the image of the transformation. IN this case the image is a plane, so every vector will be mapped to some vector within that plane

consider any vector, v. you could write it as v = k + p. where k is the component along the vector (1,1,1) and p is the rest, which will lie in the image plane

when you apply the transformation you get

T(v) = T(k+p) = A.(k+p) = A.k + A.p = 0 + A.p = A.p

and the result A.p will in the image plane

A good example to use is the projector into xy plane
[100]
[010]
[000]

the null space is any vector along the z axis, while the image is the xy plane, and in fact gives the projection of any vector onto that plane

note though that in general transformations (like yours) might rotate and stretch vectors within the image plane as well...

3. Oct 29, 2009

### lanedance

updated above

4. Oct 29, 2009

### underacheiver

For part c, why is the nullity 1? The kernel is (t,t,t) so its basis is {(1,0,0),(0,1,0),(0,0,1)}
is that not 3 dimensions?

5. Oct 29, 2009

### Staff: Mentor

Your matrix A maps any vector that is a multiple of (1, 1, 1) to the zero vector in R3. Any vector that is not a multiple of (1, 1, 1) does not get mapped to 0. A basis for this subspace is {(1, 1, 1)}.

The set of all such vectors is a one-dimensional subspace of R3, so nullity(A) = 1. Here the kernel(A) is of dimension 1, and you have two more dimensions to account for in R3.

6. Oct 29, 2009

### lanedance

As Mark pointed out {(1,0,0),(0,1,0),(0,0,1)} is not a basis for t.(1,1,1,).

And none of the vectors{(1,0,0),(0,1,0),(0,0,1)}, are in the kernel. Only the linear sum of equal amounts of each gives t(1,1,1), the null space (kernel).

Try T((1,0,0)) what do you get?

7. Oct 29, 2009

### underacheiver

but t(1,1,1) can be written as a linear combination of the set {(1,0,0),(0,1,0),(0,0,1)}.
am i getting something mixed up here?

8. Oct 29, 2009

### lanedance

(1,1,1) can be written as a linear combination of the basis you gave. This is because the basis you gave spans all of R^3, and (1,1,1) is in R^3

the difference is we want a basis for the subspace defined by the line through the origin t(1,1,1). This is teh null space of your operator

A basis is a minimal number linearly independent vectors from within the space. (1,0,0) is not within the line, hence it is not mapped to zero by the operator

when we pick a vector within the line say (2,2,2), any other vector in the line can be written as a linear multiple of (2,2,2). so (2,2,2 is a basis for the line. so is (-1,-1,-1)

9. Oct 29, 2009

### underacheiver

ah~ ok, i see. thank you.