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I want to find the change of coordinate matrix using $[I]_{\beta^{'}}^\beta$ that changes ${\beta^{'}}$ coordinates into ${\beta}$ coordinates.

$${\beta^{'}}=\{(0,10),(5,0)\}$$ $${\beta}=\{(-1,3),(2,-1)\}$$

I don' have any problems writing a linear transformation as a matrix when the basis are the standard basis but when the basis are not standard then I seem to be getting the wrong answers.

I computed $[I]_{\beta^{'}}^\beta$ by first computing the identity function with respect to $\beta^{'}$ and got

$$I(0,10)=(0,10)$$ $$I(5,0)=(5,0)$$

Then when I try to write these vectors in with respect to $\beta$, similar to how I would with respect to the standard basis I get

$$\begin{bmatrix} 30 & -5 \\ -10 & 10\\ \end{bmatrix}$$

Thanks in advance!

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  • $\begingroup$ Note that $A(A^{-1}B)=B$, with $A$ having the basis vectors of $\beta$ as columns, and $B$ having basis vectors of $\beta'$ as columns. $\endgroup$ – Abishanka Saha Dec 12 '17 at 16:15
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I'm not sure how you are getting that final matrix. The first column should be the vector $(0, 10)^T$ written in terms of the basis $\beta$ and the second column should be $(5,0)^T$ written in terms of the basis $\beta$.

In other words, to obtain the first column, you are looking to solve the following vector equation

$$ \begin{pmatrix} 0 \\ 10 \end{pmatrix} = a \begin{pmatrix} -1 \\ 3 \end{pmatrix} + b \begin{pmatrix} 2 \\ -1 \end{pmatrix} $$

This is equivalent to the system of equations

$$ \begin{align} &0 = -a + 2b \\ &10 = 3a - b \end{align} $$

Solving this I get $a = 4$ and $b = 2$ so the first column in the transition matrix is $(4,2)^T$. The second column is found analogously.

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If you can find $[I]^e_\beta$ - the change of co-ordinate transformation from $\beta$ to $e$, the standard basis - and the transformation $[I]^e_{\beta'}$ from $\beta'$ to $e$, then you can compose them, just remember that ${[I]^e_{\beta'}}^{-1} = [I]^{\beta'}_e$.

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