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I have these maps from the plane to itself where $X=(x,y)$:

$f(X):=(y,-x)$

$g(X):=(x+2y,y)$

I need to compute $fg$ and $gf$ and show that none of these compositions are simply translations or that for a point $(x,y)$, it shouldn't be mapped to $(x+a_1,y+a_2)$ for some $a_1,a_2 \in R$.

For the first composition, namely $fg$, I tried plugging in $g$ to $(x,y)$ giving $(x+2y,y)$. Then plugging these into $f$ giving $(y,-x-2y)$.

Now is this correct and enough to say that this isn't simply a translation given by the fact that x and y appear in the second term? And then could I apply this technique to the other composition, namely $gf$, for the other portion of the problem?

User123454321
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  • It's better to be a little more careful than that but it's essentially right. Show that $a_1$ and $a_2$ do not exist by assuming that they do and showing that one of them would have to be equal to two different numbers, which is a contradiction. – Matt Samuel Feb 27 '16 at 00:19
  • Okay, so once I find that $f(g(X))=(y,-x-2y)$, could I just say that we assume that $f(g(X))=(x+a_1,y+a2)$, but there is not $a_1$ such that $y$ is always equal to $(x+a_1)$ by giving two examples where $a_1$ will differ? Therefore, this isn't a translation. Would that be a slightly more rigorous proof? – User123454321 Feb 27 '16 at 00:40
  • It's enough to just say that the 2nd component of $fg$ varies with $x$, but $y + a_2$ does not. Rigor is good, but you don't have to beat the poor to death to get it. And yes, you apply the same technique for the other composition. – Paul Sinclair Feb 27 '16 at 03:38

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