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Let $f:\mathbb R^{2} \to \mathbb R^{2}$ be given by $f(x,y)=(x+y,xy)$.Then, which are correct?

  1. $f$ is surjrctive.

  2. The inverse image of each point in $\mathbb R^{2}$ under $f$ has atmost two elements.

I think $f$ is surjective. We find that $ker(f)=\{(0,0)\}$. So, $f$ is one-one and hence onto. Am I right or wrong? I have no idea about option (2).

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  • $\begingroup$ How do you conclude that $f$ is onto from the fact that $f$ is injective? $\endgroup$ – Ulrik Dec 8 '14 at 8:17
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    $\begingroup$ $f$ is not linear, so you cannot use the fact that, for a linear endomorphism in a vector space of finite dimension, $f$ is injective if and only if $f$ is surjective. $\endgroup$ – Taladris Dec 8 '14 at 8:17
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    $\begingroup$ Hint: sum and product of the roots of a quadratic equation $\endgroup$ – Taladris Dec 8 '14 at 8:20
  • $\begingroup$ choose an arbitrary point (a,b) in R^2. surjectivity means that you can find x,y such that (x+y,xy)=(a,b) for all a,b,i.e that the range is the codomain. try to work that out. don't confuse linear operators with general operators. the concepts of kernel applies to linear operators $\endgroup$ – Daniel Dec 8 '14 at 8:26
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    $\begingroup$ $f$ is not injectve ($f(1,0)=f(0,1)=(1,0)$)so your reasoning is wrong (as you've already been told, you're using results that are only valid for linear finctions, but for some reason nobody bothered to give you this counterexample to your statement). $\endgroup$ – Henrik Dec 8 '14 at 8:49
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  1. If $x+y=0$ then is it possible to have $xy>0$?

  2. If $x+y=a$ and $xy=b$ then $x(a-x)=b$. So...

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  • $\begingroup$ Hint: David's first point is not a hint at how to show $f$ is surjective. $\endgroup$ – Henrik Dec 8 '14 at 9:31
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Let's consider $(a,b)$ in the range of $f$. Then there exists $x$ and $y$ such that $x+y=a$ and $xy=b$. This means that $x$ and $y$ are roots of $P(X)=X^2-aX+b$. Since it is, a quadratic equation, there is at most two roots, so the inverse image of $(a,b)$ has at most two elements $(x,y)$ and $(y,x)$. This means that $2.$ is true.

Now, since $X^2-aX+b$ has at least one root, its disciminant is nonnegative, so $a^2-4b\geqslant 0$. So we proved that $\textrm{range}(f)\subset \{\ (a,b)\ |\ a^2\geqslant 4b\ \}$ (actually we can prove we have equality). This implies that $f$ is not surjective.

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