1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Image, Kernel

  1. Dec 6, 2015 #1
    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. jcsd
  3. Dec 6, 2015 #2

    andrewkirk

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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.
     
  4. Dec 6, 2015 #3

    Mark44

    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.

     
  5. Dec 7, 2015 #4
    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.
     
  6. Dec 7, 2015 #5
    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.
     
  7. Dec 7, 2015 #6

    Mark44

    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.
     
  8. Dec 7, 2015 #7
    so my answer is correct for Im(l), but I should have written it as:
    Im(l)= span { 2,-1 } ?

    ker(l) = (1,1)
     
  9. Dec 7, 2015 #8

    Mark44

    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>}
     
  10. Dec 7, 2015 #9
    oh ok! Thanks. So the dimension of the Kernel and Im are both = 1 (based on the hint you gave above)
     
  11. Dec 7, 2015 #10

    Mark44

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Image, Kernel
  1. Kernels and images (Replies: 2)

  2. Kernel and Image (Replies: 12)

  3. Kernel and image (Replies: 2)

  4. Kernel, Image (Replies: 1)

Loading...