For the linear transformation $T : \Bbb R^2 \to \Bbb R^2$, given by $$T(x, y) = (x + 2y, x − y)$$ find the matrix $A$ for $T$ with respect to the basis $B = \{ (1, 1), (1, −1) \}$ of $\Bbb R^2$.
Don't I need another basis?
In my book, the problems are like this:
More importantly, on a high level, what is going on here?
When your transformation goes from a space to itself (for instance: the transformation in your question goes from $\Bbb R^2$ to $\Bbb R^2$), then we can talk about transformations with respect to one basis. In particular, the problem you've led with is asking you to use the basis $B = \{(1,1),(1,-1)\}$ as a "starting basis" and as an "ending basis".
If $B = \{v_1,v_2\}$, then we describe the matrix $A$ of the transformation with respect to $B$ (sometimes denoted $[T]_B$ or $[T]_B^B$) as $$ A = \begin{bmatrix}[T(v_1)]_B & [T(v_2)]_B\end{bmatrix} $$ In your example, we have $v_1 = (1,1)$ and $v_2 = (1,-1)$, so that $$ T(v_1) = T(1,1) = (3,0) = \frac 32 \cdot (1,1) + \frac 32 \cdot (1,-1) $$ That is, we see that $T(v_1) = \frac 32 v_1 + \frac 32 v_2$. So, the coordinate vector $[T(v_1)]_B$ is given by $$ [T(v_1)]_B = \begin{bmatrix}\frac 32\\ \frac 32\end{bmatrix} $$ and this column will be the first column of our matrix $A$.