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so i have the the basises $$B=\{b_1,b_2,b_3\}$$ where: $$b_1=\begin{bmatrix} 1\\0\\0 \end{bmatrix},b_2=\begin{bmatrix} 1\\0\\1 \end{bmatrix},b_3=\begin{bmatrix} 1\\1\\1\end{bmatrix}$$

and $$C=\{c_1,c_2,c_3\}$$ where: $$c_1=\begin{bmatrix} 1\\1\\1 \end{bmatrix},c_2=\begin{bmatrix} 1\\-1\\0 \end{bmatrix},c_3=\begin{bmatrix} 1\\1\\-2\end{bmatrix}$$

And i am trying to find the change of basis matrix to $B$ to $C$ So this is supposed to be :

$$A_{B\to C}=\begin{bmatrix}[b_1]_c &[b_2]_c & [b_3]_c\end{bmatrix}$$ which after calculating the columns is: $$A= \begin{bmatrix} 1/3 & 2/3 & 1 \\ 1/2 & 1/2 & 0 \\ 1/6 & -1/6 & 0 \end{bmatrix}$$ which but this is not correct. what am i doing wrog?

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  • $\begingroup$ Are the basis vectors both specified in a common basis? $\endgroup$ – mathreadler May 20 '18 at 16:53
  • $\begingroup$ yes, both are in the standard basis $\endgroup$ – Sander May 20 '18 at 17:02
  • $\begingroup$ I agree with answerer I also think it is right according to my calculation, or if the question is interpreted the wrong way maybe it is $A^{-1}$ instead. Have you tried calculating $A^{-1}$ and checked if it is correct according to the solution? $\endgroup$ – mathreadler May 20 '18 at 17:09
  • $\begingroup$ it seems like the solution must be wrong, thanks. $\endgroup$ – Sander May 20 '18 at 17:18
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Are you sure that is not the right answer? It seems alright to me.

By definition, the "base change matrix" expresses a vector in the old basis in terms of the vectors in the new basis.

For example, with the $A$ you have computed

$$Ab_1 = \begin{bmatrix} 1/3 & 2/3 & 1 \\ 1/2 & 1/2 & 0 \\ 1/6 & -1/6 & 0 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1/3 \\ 1/2 \\ 1/6 \end{bmatrix}$$

What this means is that the "coordinates" of $b_1$ in the new basis is $(1/3,1/2,1/6)$, which should satisfy

$$b_1 = \frac 13 c_1 + \frac 12 c_2 + \frac 16 c_3$$

and it does.

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  • $\begingroup$ it seems like the solution is probably wrong. thank you $\endgroup$ – Sander May 20 '18 at 17:19

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