Given basis and example outputs, infer kernel and range of linear operator The linear operator $T : V \to \mathbb{R}^2$ is such that $T(v_1)=(−1, 1)$, $T(v_2)=(1,1)$, and $T(v_3)=(2,0)$, where $v_1$, $v_2$, $v_3$ is a basis of $V$. Compute kernel($T$) and range($T$). Is $T$ one-to-one? onto? an isomorphism?
I started by assuming $v_1 = (1, 0, 0)$, $v_2 = (0, 1, 0)$ and $v_3 = (0, 0, 1)$ so as to form a basis in $\mathbb{R}^3$.  I'm struggling to get from there to the matrix that defines $T$, from which I could calculate the kernel, range, etc.  To map from an $\mathbb{R}^3$ input to an $\mathbb{R}^2$ output, I believe T needs to be a $2 \times 3$-matrix, and my first thought is that $T$ should be $((-1, 1, 2), (1, 1, 0))$ to generate the outputs above given standard basis vectors of $\mathbb{R}^3$ as inputs.
Mostly looking for confirmation on this line of thinking.
 A: *

*dimension of $V$ is $3$ while the image space has only dimension $2$, hence it cannot be one-to-one.

*To check whether it is onto, check that $(1,1)$ and $(2,0)$ spans $\mathbb{R}^2$.
You can verify the claim by working on the particular matrix $\begin{bmatrix} -1 & 1 & 2 \\ 1 & 1 & 0\end{bmatrix}$ as what you did.
A: Your line of thinking is not entirely wrong yet it needs some corrections. Firstly, as mentioned in the comments, you cannot assume that the basis $\alpha = (v_1,v_2, v_3)$ is the canonical one, i.e. $(1,0,0),(0,1,0),(0,0,1)$. Nevertheless, even if you do not know what $v_i$ are specificaly, you can express the matrix representing the linear map $T$ with respect to the basis $\alpha$. We will denote the matrix $T_{\epsilon \alpha}$ and we introduce this notation so that we can multiply $T$ from the right with vectors expressed wrt to the $\alpha$ basis and we get the result expressed wrt to $\epsilon$ basis. 
As was also mentioned, since the map goes from $\mathbb{R}^3$ to $\mathbb{R}^2$ it cannot be injective (you simply do not have "enough" dimensions in the codomain of $T$ for the injectivity). As it seems you know how to compute kernel and image of your operator if it is represented  by a matrix, let us compute the matrix only. 
Any linear operator (let us assume only finite dimensions of domain and codomain) can be expressed by a finite matrix (dependent on the choice of basis in the domain and codomain) of which the columns are the mapped basis vectors of $\alpha$ expressed in the $\epsilon$ basis. E.g. the first column of $T_{\epsilon \alpha}$ will be $(T(v_1))_{\epsilon} = (-1,1)$ and so on for the other columns. Thus, the matrix will be
$$ 
T_{\epsilon \alpha} = 
\begin{pmatrix}
-1 & 1 & 2 \\
1 & 1 & 0 
\end{pmatrix} \ .
$$
To check that this matrix is what you are looking for, you may directly compute that it does what you expect. In other words, take $v_i$ expressed  in $\alpha$, i.e. 
$$(v_1)_{\alpha} = (1,0,0),\ \ (v_2)_{\alpha} = (0,1,0),\ \ (v_3)_{\alpha} = (0,0,1)$$
 and multiply. Let's see how this works for $v_2$
$$ 
T_{\epsilon \alpha} (v_2)_{\alpha}= 
\begin{pmatrix}
-1 & 1 & 2 \\
1 & 1 & 0 
\end{pmatrix} 
\begin{pmatrix}
0 \\
1 \\
0
\end{pmatrix} = 
\begin{pmatrix}
1 \\
1 
\end{pmatrix} = (T(v_2))_{\epsilon} 
\ .
$$
