Is $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2 ,f(x,y)=(x+4y,-3x-12y)$ surjective?

Question

I am asked to determine if $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2 ,f(x,y)=(x+4y,-3x-12y)$ is surjective, and provide a proof. I am having some trouble approaching this problem. Rewriting this as $x+4y=a$ and $-3x-12y=b$ shows that $0=-3a+b$ after adding the two equations together. I think this would mean that this function is not surjective, but I'm not sure how to approach a proof.

Answer 1

A linear algebraic approach is to note that $$f(x,y) = \begin{pmatrix} 1 & 4 \\ -3 & -12 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$ The determinant of the matrix is $(1)(-12) - (4)(-3) = -12-(-12) = 0$, and hence its image has dimension $<2$, so it cannot be a surjective map.

The purely algebraic approach that you took is also fine. You took $(a,b) \in \mathbb{R}^2$ and assumed $f(x,y)=(a,b)$ and deduced that $-3a+b=0$. So far so good. This implies that there are points in $\mathbb{R}^2$ not in the image of $f$, for example setting $a=1$ and $b=0$ gives $-3=0$, which is a contradiction; it follows that $(1,0)$ is not in the image of $f$, and hence $f$ is not surjective.

Answer 2

The function $f(x,y) = (x +4y, -3x-12y)$ is not surjective. Notice that it is difficult to get both coordinates to be negative. In particular suppose that $(-1,-1)$ is in the image of $f$. Then $$ x + 4y = -1 \qquad \text{and} \qquad -3x - 12y = -1.$$ This system of equations does not have a solution. Hence, $f$ is not surjective.

Your work was leading you down the correct path. Finding a particular counterexample is a nice way of showing a function is not surjective.