Existence of linear mapping I am studying for an exam in linear algebra and I am having trouble solving the following:
Do linear mappings $\phi : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ with the following properties exist?
$1)$  $\phi_1 \begin{pmatrix} 2 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $,   $\phi_1 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} $, $\phi_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} = \begin{pmatrix} 2 \\3 \end{pmatrix} $
$2)$  $\phi_2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 0 \\0\end{pmatrix} $,   $\phi_2 \begin{pmatrix} 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 2\\ 3 \end{pmatrix} $
$3)$  $\phi_3 \begin{pmatrix} 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 1\\ 0 \end{pmatrix} $,   $\phi_3 \begin{pmatrix} -2 \\ 5 \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix} $, $\phi_3 \begin{pmatrix} 2 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\1 \end{pmatrix} $
I know that the following properties have to hold for a linear mapping:


*

*$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y})$

*$f(\alpha \mathbf{x}) = \alpha f(\mathbf{x})$

*$f(0) = 0 $


I conclude that $2)$ is a not a linear mapping since $\phi(0)$ is not mapped to $0$.
But how shall I proceed with the others?

Hint: A linear transformation is completely determined by how it acts on a basis for your vector space (a linearly independent and spanning set), and a linear transformation can send basis vectors to wherever it wants (you can send them anywhere). Show these fact if you haven't yet!!
For $\phi_1$, note that $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$ form a basis for $\Bbb R^2$, so how $\phi_1$ acts on all of $\Bbb R^2$ if $\phi_1$ is a linear map. There is a unique linear map extending values of $\phi_1$ on this basis to all of $ \Bbb R^2$. Could $\phi_1$ be that map based on its 3rd known value?
A similar proof works for $\phi_3$.
