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Exercise counter-image with two variables

  1. Sep 9, 2016 #1
    1. The problem statement, all variables and given/known data
    I have this linear application:
    ##f : \mathbb {R}^4 \rightarrow \mathbb {R}^4##
    ##f((x, y, z, t)) = (-2x - y + t, 2x + y - t, -4x - 2y + 2t, 0)##
    and then I have this vector which I have to find the counter-image of it:
    ##\vec v = (-k, h - 2, k, h^2 - 4)##
    With ##h## and ##k## real parameters that vary.

    2. Relevant equations


    3. The attempt at a solution
    I already found out that the matrix associated with the linear application had rank ##1##.
    So, now, I put the vector in the matrix as a column and we have:
    ##(A|\vec v) = \begin{pmatrix}
    -2 & -1 & 0 & 1 & -k\\
    2 & 1 & 0 & -1 & h - 2\\
    -4 & -2 & 0 & 2 & k\\
    0 & 0 & 0 & 0 & h^2 - 4
    \end{pmatrix}##
    I know that at this time I have to find the determinants and see what happens. The problem here is that I don't know how it works with two parameters, since I've always done it with just one. Do I have two different counter-image because of this? What does it change from the single parameter exercise?
    From what I get, the determinants are ##2x2## so the rank is max ##2##, depending on the parameters. We would have:
    ##|A'| = h - 2 - k##
    ##|A''| = -2h + 4 - k##
    ##|A'''| = 2h^2 - 8##
    If the last one is ##0## (so ##h = 2##), then the rank would be the same as the rank of the linear application's matrix, and so we could proceed. But what about the ##k## in the other two cases? Should I stick with ##h = 2## and then being forced to put ##k = 0##? Or I have to do it in a different way?
     
  2. jcsd
  3. Sep 9, 2016 #2

    Mark44

    Staff: Mentor

    Find the determinants? I don't know what you mean by this.
    Just row-reduce the matrix above, and you should see what values of h and k must have.
    I don't know what the equations above represent. What are A', A'', and A'''?
    In my work, I found that it must be true that k = 0 and h = 2. The preimage of f (what you're calling the counterimage) is a three-dimensional subspace of R4.
     
  4. Sep 9, 2016 #3
    Our professor taught us to find the determinants and then the rank. If the parameter is equivalent to a number that would make the rank of the ##(A|\vec v)## equal to the rank of ##A##, then we have to find the preimage by putting the number that the parameter is associated with.
    For example, if ##h = 2## this implies that ##rank(A) = rank (A|\vec v)## and so we put all in a system and start the calculus to find the preimage.

    These are the three determinants that are different from ##0##.
    ##A' = \begin{vmatrix}
    1 & -k\\
    -1 & h - 2
    \end{vmatrix}##
    ##A'' = \begin{vmatrix}
    -1 & h - 2\\
    2 & k
    \end{vmatrix}##
    ##A''' = \begin{vmatrix}
    2 & k\\
    0 & h^2 - 4
    \end{vmatrix}##

    Should I row-reduce the matrix always in this kind of exercise?
    I would have something like this if I row-reduce:
    \begin{pmatrix}
    -2 & -1 & 0 & 1 & -k \\
    0 & 0 & 0 & 0 & h - 2 - k\\
    0 & 0 & 0 & 0 & 3k\\
    0 & 0 & 0 & 0 & h^2 - 4\\
    \end{pmatrix}
    And in a system would be:
    \begin{cases}
    -2x - y + t = -k \\
    0 = h - 2 - k \\
    0 = 3k \\
    0 = h^2 - 4
    \end{cases}
    \begin{cases}
    -2x - y + t = 0 \\
    h = 2 \\
    k = 0 \\
    h = 2
    \end{cases}
    Is this right?
     
  5. Sep 9, 2016 #4

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Forget determinants. In fact, why even bother with matrices? You have a simple linear system of 4 equations in 4 unknowns:
    $$\begin{array}{rcl}
    -2x-y+t&=&-k \\
    2x+y-t&=&h-2\\
    -4x-2y+2t&=& k\\
    0 &=&h^2-4
    \end{array}
    $$
    If ##h = 2## or ##h = -2## you will be left with three equations in the three unknowns ##x,y,t##, which you can just solve using elementary elimination methods faster than you could even write down the matrices. If ##h \neq \pm 2## your ##\vec{v}## cannot "come from" any ##(x,y,z,t)## via ##f(x,y,z,t)##. The variable ##z## does not appear anywhere in your ##f(x,y,z,t)## as you have written it.
     
    Last edited: Sep 9, 2016
  6. Sep 10, 2016 #5
    I don't understand how to resolve that system. I ended up with something like this:
    ##\begin{cases}
    -h + 2 = -k \\
    y = -2x + t + h - 2 \\
    -2h + 4 = k \\
    h = \pm 2
    \end{cases}##
    Did I do something wrong?

    I don't understand what you mean by "##\vec v## cannot come from any ##(x, y, z, t)## via ##f(x, y, z, t)##". You mean that it doesn't have a preimage if it's not ##h \neq \pm 2##?
     
  7. Sep 10, 2016 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    Yes, that is exactly what I mean: the resulting ##\vec{v}## is not in the "range" of ##f(x,y,z,t)##.

    Anyway, if ##h = 2## the equations imply that ##k = 0##, and so the resulting pre-image of ##\vec{v}## is ##\{(x,y,z,t): t = 2x+y, x,y,z,\in R \}##. Now look at the other case, ##h = -2##, to see what you get.
     
  8. Sep 10, 2016 #7
    I would have ##k = 8## and ##k = -4##, which is impossible so it HAS to be ##h = 2##, right?
     
  9. Sep 10, 2016 #8

    Ray Vickson

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    Science Advisor
    Homework Helper

    Yes, I believe so.
     
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