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Our book gives this problem:

Find the $\mathcal{B}$-matrix for the transformation $\vec{x} \rightarrow A\vec{x}$ when the basis $\mathcal{B} = \{ \vec{b}_1, \vec{b}_2 \}$, where $A = \left[\begin{array}{cc} 3 & 4 \\ -1 & -1 \\ \end{array} \right]$, $\vec{b}_1 = \left[\begin{array}{cc} 2 \\ -1 \\ \end{array} \right]$, and $\vec{b}_2 = \left[\begin{array}{cc} 1 \\ 2 \\ \end{array} \right]$.

From what I understand, it's asking us to find the matrix for the same exact transformation as $A$, except relative to to the given bases. I can't figure out where to go from here, though... any thoughts?

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up vote 2 down vote accepted

What you need to do is form the matrix $B = (\vec{b}_1|\vec{b}_2)$, where $\vec{b}_i$ is the $i$th column of B, and note that this matrix converts vectors from the standard basis into the basis $\mathcal{B}$, while the inverse $B^{-1}$ will convert vectors in the basis $\mathcal{B}$ into the standard basis. Thus if you have a vector already in the basis $\mathcal{B}$, you can convert it to standard basis by multiplying by $B^{-1}$, multiply it by $A$, and finally convert back to $\mathcal{B}$ by multiplying by $B$, so your overall matrix is $BAB^{-1}$.

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OHHHHH that's so easy! Just change coordinates, transform, and go back... wow. Thanks so much for the beautiful explanation! :) –  Mehrdad Mar 16 '11 at 8:27
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