Check if there is a real linear mapping with properties $L: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}$

Question

Check if there is a real linear mapping with properties $L: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}$

$L\begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix}=\begin{pmatrix} 1\\ 0 \end{pmatrix}$, $ \text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }L\begin{pmatrix} 0\\ 1\\ 1 \end{pmatrix}=\begin{pmatrix} 0\\ 1 \end{pmatrix}$

If there exists such linear mapping, write its matrix form.

It's from an exam from 1996, can be found here (but in German):https://moodle.hu-berlin.de/mod/resource/view.php?id=91606

I unfortunately have no idea how a task like that could be solved correctly. Maybe you can show me how it's done on this example?

I know of the conditions so a mapping is linear but how to apply them here, I don't know :s

Answer 1

Linear mapping may be represented as a matrix. Let $$ M_L = \begin{pmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{pmatrix} $$ be a matrix representating map $L$. Then we have $$ \begin{pmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{pmatrix}\begin{pmatrix} 1\\ 1\\ 1\end{pmatrix}=\begin{pmatrix} 1\\ 0\end{pmatrix}, \quad \begin{pmatrix}x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3\end{pmatrix}\begin{pmatrix} 0\\ 1\\ 1\end{pmatrix}=\begin{pmatrix} 0\\ 1\end{pmatrix} $$ or $$ \begin{cases} x_1 + x_2 + x_3 = 1 \\ x_2 + x_3 = 0 \end{cases},\quad \begin{cases} y_1 + y_2 + y_3 = 0 \\ y_2 + y_3 = 1 \end{cases}, $$ which implies $$ M_L = \begin{pmatrix}1 & x & -x \\ -1 & y & 1-y\end{pmatrix}, $$ where $x$ and $y$ are some real numbers.

Answer 2

First, assume such a matrix exists and has the form $\begin{bmatrix} a & b & c \\ d & e & f \end{bmatrix}$. If we can find the values $a,...,f$ such that the matrix operates as the function $L$, we have solved the problem.

Using the provided equalities, we get (by matrix multiplication of the vector) the conditions $a + b + c = 1$, $d+e+f=0$, $b+c = 0$ and $e+f =1$. Here we can easily solve $a = 1$ and $d = -1$.

We still have to figure out what to do with the rest of the variables. Luckily, we have more variables than constraints, so we can just decide the values for some of them. So let's try $b = 1$ and $e = 1$. This choice with the above equations gives us $c = -1$ and $f = 0$. Therefore, we get the solution $$\begin{bmatrix} 1 & 1 & -1 \\ -1 & 1 & 0 \end{bmatrix},$$ which works as the mapping $L$ that we wanted.