My book has this problem:
Consider the isomorphism $Rep_B(\cdot): P_1 \rightarrow \mathbb{R}^2$ where $B = \left\langle 1,1+x \right\rangle$. Find the image of the element $3-2x$ of the domain.
The following is my understanding of the question and thoughts on the approach: I know that an image of an function is the set of all output values. This question is saying $P_1$ is isomorphic to $\mathbb{R}^2$, which makes sense given they have the same dimensions. Then the basis for $P_1$ is $B = \left\langle 1,1+x \right\rangle$. I know I have to do something to transform the $3-2x$ into a 2-tall column vector. Can someone please elaborate on how I would use the basis to find the image?
The book answer is: $$Rep_B(3-2x)=\begin{pmatrix}5\\\ -2\end{pmatrix}$$
Thanks!
Hint: You know what $\text{Rep}_B(1)$ and $\text{Rep}_B(1 + x)$ are. Find $a, b \in \mathbb{R}$ such that $3 - 2 x = a \cdot 1 + b \cdot (1 + x)$. What do you then know about $\text{Rep}_B(3 - 2 x)$ in terms of $a$ and $b$?
Further hint: comparing coefficients yields that $$3 - 2 x = a \cdot 1 + b \cdot (1 + x) = (a+b) \cdot 1 + b x$$ for all real $x$ if and only if $$a+b = 3 \qquad b = -2,$$ which is a simple linear system you can solve.