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Thread: ordered bases

  1. #1
    Member Jskid's Avatar
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    ordered bases

    Let $\displaystyle S={\vec v_1, \vec v_2}$ and $\displaystyle T={\vec w_1, \vec w_2}$ be bases for $\displaystyle P_1$ where $\displaystyle \vec w_1 = t-1$ and $\displaystyle \vec w_2 = t+1$. If the transition matrix from T to S is $\displaystyle \left[ {\begin{array}{cc}
    1 & 2 \\
    2 & 3 \\
    \end{array} } \right]
    $ determine S.

    So the columns of the matrix are the coordinates fo the T-basis vectors with respect to the S-basis, but that's not good because S is what we're trying to find
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    There is a little problem about terminology. I don't know if your teaher means:


    $\displaystyle (i)\;[p(t)]_S=\begin{bmatrix}{1}&{2}\\{2}&{3}\end{bmatrix}[p(t)]_T\textrm{\;\;or\;\;}(ii)\;[p(t)]_T=\begin{bmatrix}{1}&{2}\\{2}&{3}\end{bmatrix}[p(t)]_S$

    Supposing $\displaystyle (i)$ we have:

    $\displaystyle [\vec{w_1}]_T=\begin{bmatrix}{1}\\{0}\end{bmatrix},\; [\vec{w_2}]_T=\begin{bmatrix}{0}\\{1}\end{bmatrix}$

    So,

    $\displaystyle [\vec{w_1}]_S=\begin{bmatrix}{1}&{2}\\{2}&{3}\end{bmatrix}\be gin{bmatrix}{1}\\{0}\end{bmatrix}=\begin{bmatrix}{ 1}\\{2}\end{bmatrix}$

    Then, $\displaystyle \vec{w_1}=t-1=\vec{v_1}+2\vec{v_2}$ etc .
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