<s>[INST] 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?
 [/INST] Very fundamental theorem says that linerar mapping from a vector space is the same thing as any function from some basis. Any such function can be uniquelly extended to whole vector space. In your case for example vectors $(2, 0)$, $(1, 1)$ are lineary independent and so form a basis of $\mathbb{R}^2$. So there exists unique linear mapping $\phi_1$ such that first two conditions hold. You have to find out if this $\phi_1$ satisfies $\phi_1(1, 2) = (2, 3)$. You can do this by decomposing $(1, 2)$ to basis $(2, 0)$, $(1, 1)$. You should be also able to write down the matrix of $\phi_1$ wrto standard basis and see what value it takes on $(1, 2)$.
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