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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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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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    $\begingroup$ OHHHHH that's so easy! Just change coordinates, transform, and go back... wow. Thanks so much for the beautiful explanation! :) $\endgroup$ – user541686 Mar 16 '11 at 8:27
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    $\begingroup$ Are you confident it's not $B^{-1}AB$? I'm convinced that it is. See 14:00 @ khanacademy.org/math/linear-algebra/alternate-bases/… $\endgroup$ – Max Li Apr 17 '18 at 5:41
  • $\begingroup$ @MaxLi Indeed it is. What they should have said is that the matrix $B$ converts the components of a vector in the $\mathcal B$ basis to its components in the standard basis. We want a matrix that transforms the components of $\vec x$ in the $\mathcal B$ basis to the components of $A\vec x$ in the $\mathcal B$ basis, so we need $B^{-1}AB$ (convert the components from $\mathcal B$ to standard, apply the transformation from standard to standard, and then convert back to $\mathcal B$). $\endgroup$ – Alex Provost Nov 15 '20 at 14:33

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