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

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.

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 $$0$$.