# Image, Kernel

1. Dec 6, 2015

### says

1. The problem statement, all variables and given/known data
Given the linear transformation l : R2 → R2 , l(x, y) = (2x − 2y, −x + y), write the matrix associated to l with respect to the standard basis of R2 , find Kerl, Im l, its bases and dimensions. Find all vectors of R2 that are mapped to (4, −2).

2. Relevant equations
Ax=0 (Nullspace / Kernel)

3. The attempt at a solution
standard basis in R2 = (1,0) and (0,1)
L(1,0) = (2,-1)
L(0,1)= (-2,1)

so matrix A =
[2 -2]
[-1 1]

If we row reduce this matrix we see that there is a free variable, which means the basis of this matrix =
[2 ]
[-1]

Dimension = 2

Im(l) =
{ [ 2] , [-2] }
[-1] , [ 1]

To get the kernel we make the matrix A =

[2 -2 | 0]
[-1 1 | 0]

Row reduction gets me to
[1 1/2 | 0]
[0 0 | 0]

ker(l) =
{ [ 1] , [1/2] }
[ 0] , [ 1 ]

[x] = X [1] - Y[1/2]
[y] = [0] [1]

Vectors mapped to (4,-2)
R2 → R2 , l(x, y) = (2x − 2y, −x + y)

[2(4) - (2)(-2), -(4)+(-2)]
= (12, -6)

I'm not sure if I've done it correctly, this is the first time I've tackled a problem like this. The fact that the ker(l) has a free variable makes me think that maybe I did something wrong -- I'm just reading up on linear transformations and linear independence / dependence now.

2. Dec 6, 2015

### andrewkirk

One can see by inspection that the kernel is generated by $(1\ \ 1)'$ (ie x=y). So something must have gone wrong in the row reduction.

3. Dec 6, 2015

### Staff: Mentor

What does "basis of the matrix" mean?
??
This can't be a basis, as the 2nd vector is the -1 multiple of the first.
Check your arithmetic. Both andrewkirk and I are getting a different vector for the basis of Ker(L).
This isn't how it works.
Once you get the matrix row-reduced, read off the equations that it represents, and use it/them to get your vectors. In this case you have an error.

4. Dec 7, 2015

### says

l (x,y) = (2x-2y, -x+y)

Ker(l): Ax = 0

[ 2 -2 | 0]
[-1 1 | 0]

row reduction

[ 1 -1 | 0]
[ 0 0 | 0]

x-y = 0, so x=y

The second column is a free / independent variable, and is the only free / independent variable in the matrix so ker(l) =
[ 1 ] , [ -1 ]
[ 0 ] [ 0 ]

I'm not sure if I've wrote this last bit correctly, I know if x=y then 1 or -1 will give me the zero vector. By writing it the way I have I'm saying in the first component of the first vector x=1 and in the second component of the second vector y=-1. Then in the second vector I'm saying both components = 0, which further proves the equality between the 2 variables.

5. Dec 7, 2015

### says

Process for finding the Im(l), which equals colum space of matrix.

A =
[ 2 -2 ] T
[ -1 1 ]

AT =
[ 2 -1 ]
[ -2 1 ]

Row reducing gives us:
[ 2 -1 ]
[ 0 0 ]

So Im(l)=
[ 2 ]
[ -1 ]

Does this also mean the Im(l) = basis of the Im(l) because there is only one vector?

I can't find any simple literature on finding the dimension. I assumed that because the transformation was from R2 to R2 the dimension just = 2.

6. Dec 7, 2015

### Staff: Mentor

x = y
y = y
The above can be written as $\begin{bmatrix} x \\ y\end{bmatrix} = k \begin{bmatrix} 1 \\ 1\end{bmatrix}$.
Geometrically, Ker(L) is the line whose equation is y = x.

No. Neither vector above is in Ker(L).
Im(L) is NOT just this vector -- it is the set of all vectors in the plane that are multiples of this vector.
Im(L) does not consist of just a single vector, but the vector you show is a basis for Im(L).
Dimension of what?
When you talk about the dimension of some subspace or vector space, say which space/subspace you're talking about.
What is dim(Ker(L))?
What is dim(Im(L))?
Hint: The number of vectors in a basis for a vector space/subspace determines the dimension of it.

7. Dec 7, 2015

### says

so my answer is correct for Im(l), but I should have written it as:
Im(l)= span { 2,-1 } ?

ker(l) = (1,1)

8. Dec 7, 2015

### Staff: Mentor

Almost.
Im(L) = span{<2, -1>}
The kernel isn't just a single vector, which is what you wrote.
Ker(L) = span{<1, 1>}

9. Dec 7, 2015

### says

oh ok! Thanks. So the dimension of the Kernel and Im are both = 1 (based on the hint you gave above)

10. Dec 7, 2015

### Staff: Mentor

Yes.
There's a theorem that you might already have seen or if not, will see shortly.
For a linear transformation $L : V \to W$, dim(Ker(L) + dim(Range(L)) = dim(V)
Here, Range(L) is essentially the same as what you're calling Im(L) -- image of L.