How to show the following function is surjective?

Question

Question:

Consider the function $f:\mathbb R^3\rightarrow \mathbb R^2$ defined by $$f(x)= \begin{pmatrix} 1&1&2\\ -1&1&0\\ \end{pmatrix}x$$ Show the function is surjective.

How can I show?

Answer 1

Note that $$ f\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}1\\-1\end{pmatrix} \qquad f\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix} $$ Note then that the vectors $\begin{pmatrix}1\\-1\end{pmatrix}$ and $\begin{pmatrix}1\\1\end{pmatrix}$ are linearly independent.

How do I know they are and what columns of the matrix should be considered? Do Gaussian elimination: $$ \begin{pmatrix} 1&1&2\\ -1&1&0\\ \end{pmatrix} \to \begin{pmatrix} 1&1&2\\ 0&2&2\\ \end{pmatrix} \qquad R_2\gets R_2+R_1 $$ The pivots are in columns 1 and 2, so the matrix has rank $2$ and a set of generators of the image is the first and second column; but rank $2$ already tells you the map is surjective.

Answer 2

Hint: Write $x = (x^1, x^2, x^3) \in \mathbb R^3$ and let any $y=(y^1,y^2) \in \mathbb R^2$ be given. The what can you say about the solutions of the following system?

$$\begin{cases}x^1 + x^2 + 2x^3 &= y^1\\-x^1 + x^2 &= y^2 \end{cases}$$

Notice that $1 \cdot 1 + 1\cdot 1 \neq 0$ ($a_1b_2- a_2b_1$), so what can you say about the configuration of those planes? What is their intersection?