# Finding vector coordinates in a basis [closed]

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!

• Calculate the transition matrix $T^{B}_{C}$. Commented Jun 9, 2019 at 14:22
• @skullph Thanks! I have perhaps a silly question. Is transition matrix from B to C the same as transition matrix from C to B? Commented Jun 9, 2019 at 14:32
• 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 $T^{B}_{C} = (T^{C}_{B})^{-1}$. Commented Jun 9, 2019 at 14:38
• @skullph So I need to do transition matrix from B to C in order to find the coordinates? Commented Jun 9, 2019 at 15:38
• Yes that should work Commented Jun 9, 2019 at 15:51

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