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How to find matrix with a given image or kernel rather than vice versa?

  1. Oct 2, 2014 #1
    I'm interested in learning how to solve a relatively general sort of problem that comes up a lot in my problem sets and will presumably come up in future exams.

    I'm asked to give an example of a matrix or linear transformation that has a given image or kernel.
    Here are some examples...

    1. The problem statement, all variables and given/known data
    Give an example of a matrix A such that im(A) is the plane with normal vector {1, 3, 2} in R^3.
    Give an example of a linear transformation hose kernel is the line spanned by {-1, 1, 2}.

    2. Relevant equations
    No equations per se.
    Some definitions?
    The image of a function consists of all the values the function takes in its target space. So if f(x) = y, y is the image.

    And the kernel of T is the solution set of the linear system Ax = 0

    3. The attempt at a solution
    For the problem with the normal vector, I've tried rotating the normal vector 90 degrees in various directions (since the plane in question is perpendicular, right?) and hoping that a two-columned matrix including two of those transformed vectors might be the answer, but I'm not even sure how to check my results.

    For the problem with the kernel, I'm just stumped how to even begin.

    These seem like relatively simple concepts, but I'm clearly missing something that I need to hurry up and just grasp. Your help would be much appreciated.
     
  2. jcsd
  3. Oct 2, 2014 #2

    WWGD

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    For 1, the set of points in the plane is given by the subspace : {##(x,y,z): x+3y-2z =0 \subset \mathbb R^3## }. Choose a basis for this subspace, extend it to a basis for ## \mathbb R^3 ##. Then map the third vector in this last basis to the
    0 vector ##(0,0,0)##. Then the linear map you want takes (1,0,0), (0,0,1) to the two basis vectors for the plane, and maps (0,0,1) to (0,0,0).

    For the kernel, define a linear map that takes (-1,1,2 ) to (0,0,0) (careful, you need to use parentheses and not just set notation, since, unlike the case of sets, the order in a basis element matters ), extend (-1,1,2) to a basis for ## \mathbb R^3 ## and send the second, third basis vectors of the extended basis to , e.g., (1,0,0), (0,1,0) ( you can use any two L.I vectors here ) in any order.

    Note that the result we are (implicitly) using here is that a bijective linear map between vector spaces is an isomorphism; this means that the linear map from the extended basis into {(1,0,0), (0,1,0)} is an isomorphism, so that it will have trivial kernel, so the whole linear map will have its kernel generated by (-1,1,2).

    To extend the basis you can use, e.g., the fundamental theorem of linear algebra.
     
    Last edited: Oct 2, 2014
  4. Oct 2, 2014 #3
    You seem to be using a lot of concepts that haven't been presented to me yet. I don't know what you mean by "bijective", linear map", "extend", "isomorphism", or the fundamental theorem of linear algebra, for example. Could you give me a short explanation of what those mean so that I can understand your advice?

    And it would seem that my biggest problem is determining a basis for {(x, y, z): x + 3y - 2z = 0}. I don't quite know how to do that.
     
    Last edited: Oct 2, 2014
  5. Oct 2, 2014 #4

    WWGD

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    O.K. , sorry I don't have that much time now, but let me give you something to start with, for defining a basis:

    You have the subspace of points satisfying x+3y-2z =0 . Solve for , say x. Then x=2z-3y . Now, set z=0 , so that
    x=-3y . Then any vector (-3y,y,0) is a basis vector. For simplicity, choose y=1 . Then you get the vector (-3,1,0) as a basis vector. For the other vector, use again x=2z-3y. This time, set y=0 . Then x=2z , and any vector (2z,0,z) is a basis vector. Again , for simplicity, set z=1, to get (2,0,1). Then the two vectors {( -3,1,0),(2,0,1)} are a basis for the plane given by { (x,y,z): x+3y-2z=0} . You can double-check by showing that these vectors are : i ) Linearly indepent, and in the plane, and ii) Every vector in the subspace can be written as a linear combination of these two.
     
  6. Oct 3, 2014 #5

    RUber

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    To provide an alternative viewpoint with essentially the same information, I would begin with (x, y, z) dot (1, 3, 2) = 0 from the definition of the normal vector. You can use this to find those basis vectors by choosing x, y, z to be anything you want (zeros and ones work well) as long as the vectors are not themselves linearly dependent.

    When I did this, I got x+3y+2z=0. Solving for x, x=-3y - 2z, letting z=0, y=1, one basis vector is (-3, 1, 0).
    Letting y=0, z=1, another basis vector may be (-2, 0, 1).
    It should be expected that there would not be a third basis, since your image is a two-dimensional plane, so the last vector is simply (0,0,0).

    To construct a matrix, simply stack the basis vectors as columns of the operator matrix A.

    In this example, A=
    -3 -2 0
    1 0 0
    0 1 0​
    To check for accuracy, you can multiply A with (x,y,z)T and take the result dotted with your normal vector to ensure this equals zero for all x,y,z in R^3.
     
  7. Oct 3, 2014 #6

    Fredrik

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    y is the image of x under the function f, not the image of f. Let's denote the domain of f by X. For each ##E\subseteq X##, the set ##f(E)=\{f(x)|x\in E\}## is said to be the image of E under f. The image of X under f is usually called the range of f, but it's very likely that this is what your book calls the image of f.

    First of all, you need to find that plane. It will contain 0, because A(0)=0. So it's a plane through the origin. That makes it a subspace. Every subspace has a basis. If you can find a basis for this one, you can find an A that sends two of the standard basis vectors of ##\mathbb R^3## to the two basis vectors you've found, and the third standard basis vector of ##\mathbb R^3## to some other point in the plane.

    It shouldn't be too hard to find an A that sends (-1,1,2) to 0. Then think about what A should look like to not send too many other vectors to 0.
     
  8. Oct 3, 2014 #7

    RUber

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    A relatively straightforward way to build a matrix A from a kernal K such that AK=0 is to use the general form: A=
    a1,1 a1,2 0
    0 a2,2 a2,3
    a3,1 0 a3,3
    Assuming, like in this example, A has dimension 3x3 and K has no zeros.
    Then, this becomes three simple relations. Plug in suitable values for your ai,j and you have a matrix with the desired properties.
     
  9. Oct 3, 2014 #8
    Thanks.
     
  10. Oct 3, 2014 #9

    Fredrik

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    A function ##f:X\to Y## is said to be

    (a) injective if the implication ##f(x)=f(y)\Rightarrow x=y## holds for all ##x,y\in X##.
    (b) surjective onto Y, or just surjective, if f(X)=Y.
    (c) bijective if it's both injective and surjective.

    "Map" is just another word for "function". If X and Y are vector spaces over ##\mathbb R## ("over ##\mathbb R##" means that the set of scalars is assumed to be the set of real numbers, rather than some other field, like the set of complex numbers), then a map ##T:X\to Y## is said to be linear if ##T(ax+by)=aT(x)+bT(y)## for all ##x,y\in\mathbb R##.

    If V is a vector space and ##S\subseteq V##, to extend S to a basis for V is to find a set S' that has S as a subset and is a basis for V.

    Find a vector in that plane. Then find a vector in that plane that isn't a multiple of the first vector you found.
     
  11. Jan 31, 2015 #10
    One Simple recipe to acomplish this is the following:

    say the kernel is <(0,0,1),(1,2,3)>
    and the image is <(1,2,3)> (i.e. the line spanned by the vector (1,2,3))

    which of course means both (0,0,1) and (1,2,3) get mapped to (0,0,0)
    and _some_other_ vector gets mapped to (1,2,3), our arbitrary choice is (1,0,0)
    of course there is no unique solution, we just need to find a candidate.

    So we write the vectors in rows on the left, and their corresponding images on the right
    $$
    \left[\begin{array}{rrr|rrr}
    0& 0 & 1 & 0 & 0 & 0 \\
    1& 2 & 3 & 0 & 0 & 0 \\
    1& 0 & 0 & 1 & 2 & 3
    \end{array}\right]
    $$

    solve it with gauss:
    $$
    \left[\begin{array}{rrr|rrr}
    1& 0 & 0 & 1 & 2 & 3 \\
    0& 1 & 0 & -1/2 & -1 & -3/2 \\
    0& 0 & 1 & 0 & 0 & 0
    \end{array}\right]
    $$

    Transpose the solution - the matrix you were trying to find is:
    $$
    \begin{pmatrix}
    1 & -1/2 & 0 \\
    2 & -1 & 0 \\
    3 & -3/2 & 0
    \end{pmatrix}
    $$

    Basically we just inverted the process you would use to find the image of a given matrix.
     
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