Change of basis of a linear map from $\mathbb{R}^2→\mathbb{R}^3$ I have the following question in my workbook:
The linear map      $f:\mathbb{R}^2→\mathbb{R}^3$ is given by
$f(x,y)=(−2x+y,−x−y,−x−3y)$
Consider the following new bases:

*

*in $\mathbb{R}^2: [1, 2], [1, 1]$


*in the image: $[0, −3, −7] , [3, 9, 19]$
Provide the matrix of $f$ with respect to these new bases.
This question is really throwing me off, any advice on how to approach this would be greatly appreciated.
 A: So to find the matrix, we need to investigate how the linear map acts on the basis elements, and represent that in our new basis. In particular, we have the two bases $\mathcal{B}=\{(1,2),(1,1)\}$ and $\mathcal{B}'=\{(0,-3,-7),(3,9,19)\}$ (which we have ordered), and want to express $f(v)$ in $\mathcal{B}'$ for each $v\in\mathcal{B}$. Let's write
$$v_1=(1,2),\quad v_2=(1,1),$$
$$u_1=(0,-3,-7),\quad u_2=(3,9,19).$$
Now as
$$f(x,y)=(-2x+y,-x-y,-x-3y),$$
we get that
$$f(v_1)=u_1,\quad f(v_2)=-\frac{1}{3}u_1-\frac{1}{3}u_2$$
(I'll leave the details to you). We thus get the matrix of $f$ with respect to $\mathcal{B}$ and $\mathcal{B}'$ as
$$\begin{pmatrix}
1&-\frac{1}{3}\\
0&-\frac{1}{3}
\end{pmatrix}.$$
In particular, the columns are the coordinates of the images of the basis vectors of $\mathcal{B}$ with respect to $\mathcal{B}'$. The reason this is always the case is that the matrix is supposed to encompass how the linear map maps coordinates in the first basis to coordinates in the second basis, which is precisely what will occur if this is the case (I advise you play around with it a bit if it's not immediately clear why).
