So I'm having trouble finding a pattern when dealing with these types of questions; I need to find a better way to solve them: Here's the one i'm currently dealing with:
Find the range space and rank of the map: $a) f: \mathbb R^2 → P^3$ given by $(x,y) --> (0, x-y, 3y) $ (these are vectors btw)
So I get how to find range space/rank but the answer shows that "any vector $(0, a, b)$ is the image under $f$ of this domain vector:
$(a+b/3, b/3) $ <----how do i get this?? what is that and how do they find it?? I tried setting up a matrix to solve, idk what to do.
Since your map is linear, you can write it in matrix notation as follows: $$ f(x, y) = \begin{pmatrix}0 & 0\\ 1 & -1 \\ 0 & 3\end{pmatrix} \begin{pmatrix}x\\y\end{pmatrix} $$ The rank of this matrix is 2. You can also notice that it maps your 2D vector onto a plane in $\mathbb{R}^3$. Moreover, from the definition of $f(x, y)$ it follows that the first coordinate of the map is going to be always 0. Hence, the range of $f(x, y)$ is a plane in $\mathbb{R}^3$ each point of which will have coordinates of the form $(0, a, b)$. So, the answer that you quoted is correct.
Here is an example. Consider an arbitrary point $(0, a, b)$. The claim "it is the image under $f$" simply means that there exist such vector $(x, y)$ in the original space, $\mathbb{R}^2$, so that $f(x, y) = (0, a, b)$. You can find out that $(x, y) = (a + b/3, b/3)$ just by solving the following system of equations: $$ x - y = a, \quad 3 y = b $$ This gives the answer you were confused about.