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Linear transformation from given matrices - bases unknown

  1. Jul 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Find a linear transformation

    [tex]T: P_2 \rightarrow M_{22}[/tex]

    such that

    [tex]T(1+x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right][/tex]

    [tex]T(x+x^2)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right][/tex]

    [tex]T(1+x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]


    3. The attempt at a solution

    In all the examples I have access to the linear transformation is already defined, the bases given and the question is to find the matrix (easy enough).

    For this question, I could've reversed this approach had I known what the bases were. However, the bases aren't given and I'm stumped.

    Could someone offer me a hint here please?

    Thanks!
    phyz
     
  2. jcsd
  3. Jul 21, 2009 #2

    Dick

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    T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1). If you knew what T(x^2), T(x) and T(1) were it would be easy to write down a formula for T acting on a general element of P2. Can you find them from the given information?
     
  4. Jul 22, 2009 #3
    Hi Dick!

    I think I know where you're going, but not sure how to get there. Let's see if I understand what you're saying though and give it a shot:

    From what you've given me
    [tex]T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)[/tex]

    I thought that we could then say

    [tex]T(1)+T(x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right][/tex]

    [tex]T(x)+T(x^2)=\left[\begin{array}{cc}0&1\\1&0\end{array}\right][/tex]

    [tex]T(1)+T(x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

    However, I think we'll still be acting legit when swapping rows 1 and 2 of the second equation so that

    [tex]T(1)+T(x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right][/tex]

    [tex]T(x)+T(x^2)=\left[\begin{array}{cc}1&0\\0&1\end{array}\right][/tex]

    [tex]T(1)+T(x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

    If we then subtract equation 3 from 1, we get

    [tex]T(1)+T(x)-Y(1)-T(x^2)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right]-\left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

    [tex]T(x)-T(x^2)=\left[\begin{array}{cc}1&0\\0&-1\end{array}\right][/tex]

    which the same as equation 2 except for the minus sign. If all of what I've done so far is still ok and if my reasoning still holds, this means that

    [tex]T(1)=\left[\begin{array}{cc}0&0\\0&0\end{array}\right][/tex]

    [tex]T(x)=\left[\begin{array}{cc}1&0\\0&0\end{array}\right][/tex]

    [tex]T(x^2)=\left[\begin{array}{cc}0&0\\0&1\end{array}\right][/tex]

    Having done all of this I have to admit that
    1. I don't know if what I've done is mathematically acceptable and
    2. if everything up until this point is ok, what do I do next?

    Thanks for the help!
     
  5. Jul 22, 2009 #4

    Dick

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    It's perfectly mathematically acceptable. Now just add things up T(ax^2+bx+c)=aT(x^2)+bT(x)+c=?? What's the resulting matrix?
     
  6. Jul 29, 2009 #5
    Hi Dick!

    Sorry about the slow response time, but I've been rather busy this past week.

    I believe our final answer is:


    [tex]T(ax^2+bx+c)=aT(x^2)+bT(x)+cT(1)[/tex]

    [tex]=a\left[\begin{array}{cc}0&0\\0&1\end{array}\right]+b\left[\begin{array}{cc}1&0\\0&0\end{array}\right]+c\left[\begin{array}{cc}0&0\\0&0\end{array}\right][/tex]

    [tex]=\left[\begin{array}{cc}b&0\\0&a\end{array}\right][/tex]

    But I'm not so sure about the notation, do we write it as

    [tex]\left[T(p(x))\right]=\left[\begin{array}{cc}b&0\\0&a\end{array}\right][/tex]

    ?
     
  7. Jul 29, 2009 #6

    Dick

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    How about just writing T(ax^2+bx+c)? I think that's a little clearer.
     
  8. Jul 29, 2009 #7
    Ok cool. Once again thank you for your help! :smile:
     
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