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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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If I understand correctly, this is not quite what I was looking for. Let's take my exercise as an example. For the inverse function, we have the transformation matrix \begin{pmatrix} 1 & -1 & 4\\ 0 & 1 & -4\\ 0 & 0 & 1 \end{pmatrix} From this I can see that $F^{-1}(1) = 1, F^{-1}(x) = x-1, F^{-1}(x^2) = x^2-4x+4$ and due to linearity (is the inverse map linear too?), I can find $F^{-1}(s(t))$ for all $s(t) \in \mathbb{R}_2[t]$. However, I'm looking for a "more" explicit expression, something like $F^{-1}: s(t) \mapsto ...$, if you understand me... –  Huy Jun 29 '11 at 20:17
    
@Huy: Yes, the inverse of an invertible linear transformation is necessarily a linear transformation. As for a formula, you can obtain it from linearity: $F^{-1}(a+bx+cx^2) = aF^{-1}(1) + bF^{-1}(x) + cF^{-1}(x^2) = (a-b+4c) + (b-4c)x + cx^4$. If you look carefully, the coefficients of the basis vectors in the image can be "read off" the rows of the inverse matrix. –  Arturo Magidin Jun 29 '11 at 20:22
    
Because the coordinate vector of $a+bx+cx^2$ relative to the basis $[1,x,x^2]$ is $(a,b,c)^t$, so the image would be $$\left(\begin{array}{rrr}1 & -1 & 4\\0 & 1 & -4\\0&0&1\end{array}\right)\left(\begin{array}{c}a\\b\\c\end{array}\right)= a\left(\begin{array}{c}1\\0\\0\end{array}\right) + b\left(\begin{array}{r}-1\\1\\0\end{array}\right) + c\left(\begin{array}{r}4\\-4\\1\end{array}\right),$$and what's on the top line is the coefficient corresponding to $1$ (the first basis vector), what's on the middle line is the coefficient of $x$, and what's on the third line is the coefficient of $x^2$. –  Arturo Magidin Jun 29 '11 at 20:40
    
@Huy: Sorry: I forgot the ping and it's too late to edit. Hopefully you'll see the reply now. –  Arturo Magidin Jun 29 '11 at 20:46
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