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.
Asked
Active
Viewed 46 times
0
-
Sorry, should be $0=-3a+b$. By multiplying the first equation by -3 and and adding the two equations together. – Brandon O'Neil Feb 08 '18 at 02:45
-
How much Linear Algebra are you allowed to use? The easiest way is to write $f$ as a matrix and check the determinant is nonzero. – max_zorn Feb 08 '18 at 02:46
-
Still not quite right, alas. – Angina Seng Feb 08 '18 at 02:46
-
It should be $3a+b=0$ – Prasun Biswas Feb 08 '18 at 02:46
-
Basically, we have $$f(x,y)=(x+4y,-3x-12y)=\bigl(x+4y,-3(x+4y)\bigr)=(x+4y)\cdot(1,-3)$$ which shows that the image is a 1-dimensional subspace of $\Bbb R^2$, hence not surjective. – Prasun Biswas Feb 08 '18 at 02:48
-
Solve for example $,f(x,y)=(0,1),$ for $,x,y,$. If no solutions exist, then $f$ is not surjective. – dxiv Feb 08 '18 at 02:49
2 Answers
0
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.
Bryan Curtis
- 228
- 1
- 6
0
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.
Clive Newstead
- 64,925