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Finding the matrix of $T(p(x)) = p(2x-1)$ with respect to the basis $B = $ {$1+x, 1-x, x^2$}

To find the matix of a transformation with respect to a given basis, I find the images of the basis vectors under the transformation, then use the columns of those images with respect to $R^n$ as the columns of the matrix I'm looking for. Here's what I mean:

$T(1+x) = 1 + (2x - 1) = 2x$

$T(1-x) = 1 - (2x - 1) = 2 - 2x$

$T(x^2) = 1 - (2x - 1)^2 = 1 - 4x + 4x^2 $

With respect to the basis {${e_1, e_2, e_3}$} of $R^3$, those vectors would be:

$[0,2,0], [2,-2,0],[1,-4,4]$

respectively.

And so the matrix of $T$ with respect to $B$ is:

$$ \left[ \begin{array}{ccc} 0&2&1\\ 2&-2&-4\\ 0&0&4 \end{array} \right] $$

This is apparently completely wrong. The correct matrix is:

$$ \left[ \begin{array}{ccc} 1&0&-3/2\\ -1&2&5/2\\ 0&0&4 \end{array} \right] $$

My approach has been working thus far, but here it fails. Is there something different about this example that makes it a situation where I can not apply my strategy? What exactly is it that I found (if anything)?

Any help at all is greatly appreciated.

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You should express the image of each vector of $B$ as a linear combination of elements of $B$. For instance\begin{align}T(1+x)&=2x\\&=(1+x)-(1-x)\\&=1\times(1+x)+(-1)\times(1-x)+0\times x^2.\end{align}That's why the entries of the first column of the matrix are $1$, $-1$, and $0$.

Can you take it from here?

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