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Quick Linear Algebra Question

  1. Oct 12, 2006 #1
    [tex] \left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right) [/tex].


    How did you get rid of the parameter [tex] \alpha [/tex] to transform it into an equation?

    thanks
     
    Last edited: Oct 12, 2006
  2. jcsd
  3. Oct 12, 2006 #2

    quasar987

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    could you rephrase the question plz?

    transform what into an equation?
     
  4. Oct 12, 2006 #3
    I know that [tex] \left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right) [/tex] is equivalent to [tex] y = -x+2 [/tex]. How do we get this? The first matrix, I know, is the point (2,0). The second matrix is a direction vector, but I dont know where to go from there.
     
  5. Oct 12, 2006 #4

    Hurkyl

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    Ah, I see. He's starting with a parametric equation for a line:

    [tex]\left(\begin{array}{c}x \\ y\end{array}\right) =
    \left(\begin{array}{cc} 2-\alpha\\\alpha \end{array}\right) =\left(\begin{array}{cc} 2\\0\end{array}\right) +\alpha\left(\begin{array}{cc} -1\\1\end{array}\right) [/tex]

    and he wants to know how to turn it into an implicit form. (i.e. the solutions to an equation)


    There's a simple approach that works: just take the equations

    [tex]x = 2 - \alpha[/tex]

    [tex]y = \alpha[/tex]

    and eliminate the variable [itex]\alpha[/itex].


    There's a highbrow approach too, that (IMHO) is a good example to learn.

    To simplify things, let me write the original equation as:

    [tex]\vec{z} = \vec{b} + \alpha \vec{v}[/tex]

    In order to get rid of [itex]\alpha[/itex], you need to multiply by something that kills off [itex]\vec{v}[/itex]: you want a good matrix [itex]A[/itex] such that [itex]A \vec{v} = 0[/itex]. Well, that's just a left-nullspace matrix computation! (i.e. the columns of [itex]A^T[/itex] are a basis for the nullspace of [itex]\vec{v}^T[/itex])

    In this case, we get something like

    [tex]A = \left(\begin{array}{cc}1 & 1\end{array}\right)[/tex]

    And so the desired equation is

    [tex]A \vec{z} = A \vec{b}[/tex]

    which you can check reduces to [itex]x + y = 2[/itex].
     
    Last edited: Oct 12, 2006
  6. Oct 12, 2006 #5
    thank you Hurkyl.
     
  7. Oct 12, 2006 #6

    Hurkyl

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    P.S. why did I kill off [itex]\vec{v}[/itex] by using a matrix such that [itex]A\vec{v} = 0[/itex], rather than using a matrix such that [itex]\vec{v}A = 0[/itex]?
     
  8. Oct 12, 2006 #7
    because multiplication of matrices is not commutative?
     
  9. Oct 12, 2006 #8

    Hurkyl

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    Noncommutativity is why the choice matters.

    I can kill off the [itex]\vec{v}[/itex] term by finding a matrix [itex]A[/itex] such that [itex]\vec{v}A = 0[/itex], and then getting the equation

    [tex]\vec{z} A = \vec{b} A[/itex]

    but for this particular problem, doing this is bad... whereas doing it the other way is good. Can you tell why?


    Edit: bleh, the full answer isn't as straightforward as I thought. Sorry about that. :frown: It's easy to see why the method in this post won't work (A is degenerate: it has zero columns), but I'm too tired to explain why multiplying on the left works, and what to do in the more general case of something like:

    [tex]Z = B + \alpha M[/tex]

    where Z, B, and M are all mxn matrices, rather than just vectors.
     
    Last edited: Oct 12, 2006
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