Find the Inverse of a Linear Transformation - Tips I would like to know some tricks to compute $T^{-1}$.
For example, given $T: \mathbb{R}^2 \to \mathbb{R}_2[x]$ Linear Transformation, such that:
$$T((1,0))=1+x,~~~T((1,1))=1-x$$
Find $T^{-1}(2x)$.
At first, I was trying to compute $T$ itself, which is $T((a,b))=a+(a-2b)x$ and then I was trying to compute $T^{-1}$ with no luck.
After some help, I saw that $T^{-1}(2x)=T^{-1}((1+x)-(1-x))$ and that was very easy to compute.
Unfortunately I didn't notice that at first, so I would like to know how you would approach questions like this one.
Thanks a lot!
 A: In case you wanted to go along your original method, there's always the brute-force method of finding the matrix of $T$ w.r.t. the standard basis of $\mathbb{R}^2$ and the basis $\{1,x\}$ of the codomain, then inverting the matrix. So here you'd get $T(a,b)=\begin{bmatrix} 1 & 0 \\ 1 & -2\end{bmatrix}\begin{bmatrix}a \\ b\end{bmatrix}$, since this way $\begin{bmatrix}1 & x\end{bmatrix}T(a,b)=a + (a-2b)x$.
Then $T^{-1}$ is given by the matrix $$\frac 1 2\begin{bmatrix} 2 & 0 \\ 1 & -1\end{bmatrix}$$
with respect to the bases $\{1,x\}$ and the standard basis of $\mathbb{R}^2$. And indeed,
$$T^{-1}(2x) = \frac 1 2\begin{bmatrix} 2 & 0 \\ 1 & -1\end{bmatrix}\begin{bmatrix}0\\2\end{bmatrix}=\begin{bmatrix}0\\ -1\end{bmatrix}$$
But in general it's worth it to do everything you can to avoid using this approach, since it's nice to avoid having to compute matrix inverses. So, like you found in your case, it's nice to try to see if you can easily write $2x$ as a linear combination of the given values of $T(a,b)$ that you're given before you proceed to the full matrix computation.
