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Linear Algebra (Similar Matrices)

  1. Jul 20, 2009 #1
    1. The problem statement, all variables and given/known data

    Write the standard matrix representation TE for T.

    2. Relevant equations


    3. The attempt at a solution

    a+b-c \\
    c-d \\
    a-b \\

    I think this is right, but a subsequent problem asks me to show that this matrix is similar to another matrix. Similarity only makes sense for an n x n matrix, so this must be wrong.

    Any ideas?
  2. jcsd
  3. Jul 20, 2009 #2


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    The basis 'vectors' are x^3, x^2, x and 1. So ax^3+bx^2+cx+d=(a,b,c,d).This is exactly the same as other exercises you have already done, it's T(a,b,c,d)=(0,a-b,c-d,a+b-c). Sure, it's an 4x4 matrix.
  4. Jul 20, 2009 #3


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    Is T here a function from the space of polynomials of degree 3 or less to itself or to the space of polynomials of degree 2 or less? If the former, the matrix is 4 by 4. If the latter, it is 4 by 3 (but the first row is all "0"s).
    For example, taking x3 as the first "basis vector", T(x3)= ax2+ 0x+ a as a polynomial of degree 2 or less or 0x3+ ax2+ 0 x+ a as a polynomial of degree 3 or less. That is:
    [tex]T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}a \\ 0 \\ a\end{bmatrix}[/tex]
    in the first case or
    [tex]T\begin{bmatrix}1 \\ 0 \\ 0 \\ 0\end{bmatrix}= \begin{bmatrix}0 \\ a \\ 0 \\ a\end{bmatrix}[/tex]
    in the second.

    Do you see that those give you the first column of the matrix?
  5. Jul 20, 2009 #4
    Ah yes, I see. I was thinking of T as a vector instead of a transformation. Thanks!
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