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We have two bases: $B =$ {$ (2,6,-1), (3,3,1), (1,0,1)$} and $C =$ {$ (2,0,0), (1,1,2), (1,-1,-8)$}. I'm trying to find the coordinates of a vector $x$ in the basis $C$, if $x_B = (2,1,-3)$ .

What steps should I do in order to calculate the coordinates? Thanks!

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    $\begingroup$ Calculate the transition matrix $T^{B}_{C}$. $\endgroup$
    – n7kvz
    Commented Jun 9, 2019 at 14:22
  • $\begingroup$ @skullph Thanks! I have perhaps a silly question. Is transition matrix from B to C the same as transition matrix from C to B? $\endgroup$
    – james F.
    Commented Jun 9, 2019 at 14:32
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    $\begingroup$ No quite, the transition matrix from $B$ to $C$ gets you the coordinates of your basis vectors from $B$ in the basis $C$ whereas the transition matrix from $C$ to $B$ gets you the coordinates of your basis vectors from $C$ in the basis $B$. If you have $T^{C}_{B}$ however, you can get $T^{B}_{C}$ by $T^{B}_{C} = (T^{C}_{B})^{-1}$. $\endgroup$
    – n7kvz
    Commented Jun 9, 2019 at 14:38
  • $\begingroup$ @skullph So I need to do transition matrix from B to C in order to find the coordinates? $\endgroup$
    – james F.
    Commented Jun 9, 2019 at 15:38
  • $\begingroup$ Yes that should work $\endgroup$
    – n7kvz
    Commented Jun 9, 2019 at 15:51

1 Answer 1

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You are looking for $(\alpha,\beta,\gamma)$ such that $ \alpha(2,0,0)+\beta(1,1,2)+\gamma(1,-1,8)=2(2,6,-1)+(3,3,1)-3(1,0,1)$

i.e. $\begin{pmatrix} 2 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 2 & 8 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix}=\begin{pmatrix} 2 & 3 & 1 \\ 6 & 3 & 0 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$

$\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix}=\begin{pmatrix} 2 & 1 & 1 \\ 0 & 1 & -1 \\ 0 & 2 & 8 \end{pmatrix} ^{-1}\begin{pmatrix} 2 & 3 & 1 \\ 6 & 3 & 0 \\ -1 & 1 & 1 \end{pmatrix} \begin{pmatrix} 2 \\ 1 \\ -3 \end{pmatrix}$

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