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?
 A: You're quite right about $\phi_2$. For the others, you should see whether you can compute what one of the three values ought to be just using the two given values. So for instance $\begin{pmatrix}2\\1\end{pmatrix}=4\begin{pmatrix}1\\-1\end{pmatrix}+\begin{pmatrix}-2\\5\end{pmatrix}$. Can you conclude anything from this observation about whether the proposed value for $\phi_3\begin{pmatrix}2\\1\end{pmatrix}$ is consistent with the first two givens? If it is consistent, then you just need to argue that there's no inconsistency in the first two points. If it's not, then no such map can exist.
A: 1) You are given the image of $3$ vectors whereas you only need $2$ to define the mapping on $\Bbb R^2$ (because $\Bbb R^2$ is of dimension $2$). So you can assume you defined $\phi$ with the first two and then check if the $3^\text{rd}$ one is true. Now $\begin{pmatrix}1\\2\end{pmatrix}=2\begin{pmatrix}1\\1\end{pmatrix}-\cfrac{1}{2}\begin{pmatrix}2\\0\end{pmatrix}$ so we should have $\begin{pmatrix}0\\1\end{pmatrix}=2\begin{pmatrix}5\\2\end{pmatrix}-\cfrac{1}{2}\begin{pmatrix}2\\3\end{pmatrix}$ which is not the case so the answer is No.
2) As you said, $0$ should be mapped to $0$, and it is not so No.
3) It's the same thing as 1), you need the image of two vectors foming a basis to define your mapping but you have $3$ so you have to check whether the $3^{\text{rd}}$ one is consistent with the two others.
A: 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 $(1, 2)$.
A: 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 $\phi_3$.
