Ex 1

Let B & C be bases of a vector space
Given a vector $\vec{x} \text{ and } \begin{bmatrix} \vec{x} \end{bmatrix}_B$ how do we find $\begin{bmatrix} \vec{x} \end{bmatrix}_c$?

B = {$\vec{b}_1,\vec{b}_2$} and C = {$\vec{c}_1, \vec{c}_2$}
such that
$\vec{b}_1 = 2\vec{c}_1 - \vec{c}_2$ and $\vec{b}_2 = \vec{c}_1 + 3\vec{c}_2$

Let $\vec{x} = \vec{b}_1 - 8\vec{b}_2$

$\begin{bmatrix} \vec{x} \end{bmatrix}_c = \begin{bmatrix} \vec{b}_1 - 8\vec{b}_2 \end{bmatrix}_c = \begin{bmatrix} \vec{b}_1 \end{bmatrix}_c - 8 \begin{bmatrix} \vec{b}_2 \end{bmatrix}_c = \begin{bmatrix} 2\\-1 \end{bmatrix} - 8\begin{bmatrix} 1\\3 \end{bmatrix}$

$\begin{bmatrix} \vec{x} \end{bmatrix}_c = \begin{bmatrix} -6\\-25 \end{bmatrix}$

Put in a more complex, yet more compact way, in general

$\begin{bmatrix} \vec{x} \end{bmatrix}_c = P_{C <-- B} \begin{bmatrix} \vec{x} \end{bmatrix}_B$


$P_{C <-- B} = \begin{bmatrix} \begin{bmatrix} \vec{b}_1 \end{bmatrix}_c, \begin{bmatrix} \vec{b}_2 \end{bmatrix}_c,...,\begin{bmatrix} \vec{b}_n \end{bmatrix}_c \end{bmatrix}$

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