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Homework Help: Matrix as a transformation?

  1. Jul 9, 2011 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations

    3. The attempt at a solution

    I'm completely lost on this one.

    I think the question is saying matrix A is a representation of a linear transformation, with the a11 and a21 transforming to a31, and so on for the other two columns. But I don't see how you can get that result with a12 being negative. This is all I could come up with after days of looking at this problem, and I'm just going around in circles :frown:

    As for the product C=BA well that's pretty straightforward.

    If someone can point me in the right direction it will be very much appreciated!
  2. jcsd
  3. Jul 9, 2011 #2


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    so consider A as something like
    [tex] A = \begin{pmatrix} M & \textbf{0} \\ \textbf{0} & 1 \end{pmatrix} [/tex]

    [tex] M = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} [/tex]

    and consider the product
    [tex] M = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} [/tex]

    now relate that back to the original matrix A and its operation on and arbitrary x1,x2 and x3
  4. Jul 9, 2011 #3
    So when you say the product..

    [tex] M = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} [/tex]

    Is x1 and x2 equal to m11 and m12?

    Which multiplied by the matrix M = [itex]\left[\stackrel{1}{0}\right][/itex]

    Edit:- I'm still not getting it. Matrices aren't making a lot of sense to me right now :(
    Last edited: Jul 9, 2011
  5. Jul 9, 2011 #4


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    I think lanedance spoke wrong in the last part of his response. I am sure he meant to say
    [tex]A\begin{pmatrix}x_1 \\ x_2\\ x_3\end{pmatrix}= \begin{pmatrix}M & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}[/tex]

    [tex]y_1= \begin{pmatrix}x_1 \\ x_2\end{pmatrix}[/tex]
    and [itex]y_2= x_3[/itex].

    Of course, M is now a rotation matrix.
  6. Jul 9, 2011 #5


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    yeah thanks Halls, i didn't mean to have that last equals sign
  7. Jul 9, 2011 #6
    Ok.. So for A * (x1,x1,x3), x1,x2 are mapped to new points, and x3 remains at the same point. Or with A partitioned to the 2x2 matrix M, and M * (x1,x2) + (0,0,x3) * A, you get exactly the same result right..

    Is the partitioning completely arbitrary? I think that's what has been confusing me this whole time
  8. Jul 9, 2011 #7

    [tex]\begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}\begin{pmatrix} x_1 \\ x_2 \end{pmatrix} [/tex]
  9. Jul 10, 2011 #8


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    The only reason to look at the partitioning is that x3 is unchanged by the operation in this problem.

    So to understand the action of the operator we only need to consider its action on x1 and x2, which is a rotation in that plane, hence why we consider the partitioned matrix.
  10. Jul 10, 2011 #9
    Ah.. I think it finally makes sense to me now. Thank you!
  11. Jul 10, 2011 #10


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    no worries ;)
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