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I know pretty well how to find the transformation matrix of a linear map (with respect to a basis). However, I am wondering whether it is also possible to do it the other way around?

This question arises because in one of my exercises in linear-algebra I first had to find the transformation matrix with respect to the canonical basis to the linear map F: \mathbb{R}_2[t] \to \mathbb{R}_2[t],\; s(t) \mapsto s(t) + s'(t) + ts''(t), then I have to find the inverse map $F^{-1}$. The sample solution only provides the inverse of the transformation matrix, however, I want to know whether I can give an explicit function (using the inverse of the transformation matrix).

Thanks for any help in advance!

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Sure: if you have the matrix $A$ relative to the basis $\beta$, then the linear transformation that maps the coordinate vector $[\mathbf{v}]_{\beta}$ of $\mathbf{v}$ with respect to $\beta$ to $[A\mathbf{v}]_{\beta}$ has coordinate matrix $A$ with respect to $\beta$.

Or to be more explicit, if $A = \left(\begin{array}{ccc} a_{11} & \cdots & a_{1n}\\ a_{21} & \cdots & a_{2n}\\ \vdots & \ddots & \vdots\\ a_{n1} & \cdots & a_{nn} \end{array}\right),$ is the matrix relative to the basis $\beta=[\mathbf{v}_1,\ldots,\mathbf{v}_n]$, then the linear transformation defined by $T(\mathbf{v}_j) = a_{1j}\mathbf{v}_1 + a_{2j}\mathbf{v}_2 + \cdots +a_{nj}\mathbf{v}_n$ has coordinate matrix $A$ with respect to $\beta$.

In short: just "read off" what the linear transformation does to the $j$th vector of the basis by reading down the $j$th column of $A$.

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    @Huy: Sorry: I forgot the ping and it's too late to edit. Hopefully you'll see the reply now.2011-06-29