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Hey so I got a question about Vector spaces

Let $V=(8,\infty)$. For $u,v$ in $V$ and $a$ in $\mathbb R$ define vector addition by $u\boxplus v:= uv-8(u+v)+72$ and scalar multiplication by $a\boxdot u :=(u-8)^a +8$. It can be shown that $(V,\boxplus,\boxdot)$ is a vector space over the scalar field $\mathbb R$. Find the additive inverse of 16.

So what I did was find the zero vector which is 9, and set u to 16, but that was the incorrect answer. So I'm confused as to how you find the additive inverse.

Thanks!

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  • $\begingroup$ $16\boxplus x = 9 \Leftrightarrow 16x - 128 - 8x + 72 = 9 \Leftrightarrow 8x = 65 \Leftrightarrow x = 1 + \dfrac1{8}$. You can also put $a=-1$ in the definition of scalar multiplication and obtain the same result. $\endgroup$ Commented Apr 27, 2014 at 18:44
  • $\begingroup$ Ah, I think you mean 8+ 1/8 but yes, thank you! $\endgroup$
    – Kevin
    Commented Apr 27, 2014 at 18:54
  • $\begingroup$ @FPE You could make that an answer... :) $\endgroup$
    – apnorton
    Commented Apr 27, 2014 at 18:55

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Hint $\ $ The vector space arises by transport of structure from the vector space $\,(0,\infty)\,$ to $\,(8,\infty)\,$ along the shift-bijection $\,s(v)= v+8,\,$ with inverse $\,t(v) = v-8.\,$ The former vector space has the simpler vector operations given by $\ a u \oplus b v = u^a v^b.\,$ Transporting is defined so that the the maps $\,s,t\,$ are vector space isomorphisms, e.g. $\ t(u\boxplus v) = t(u)\oplus t(b)\,$ so the transported addition is

$$ u\boxplus v\, =\, t^{-1}(t(u)\oplus t(v))\, =\, (u\!-\!8)(v\!-\!8)+8\, =\, uv -8(u\!+\!v)+ 72$$

Thus you can compute inverses straightforwardly by first performing the simpler inverse computation in the space $\,(0,\infty)\,$ then shift-transporting the solution to $\,(8,\infty),\,$ as above.

Remark $\ $ Although this problem is so simple that one could skip the conceptual transportation viewpoint and do the computation by brute-force, this would be much more difficult to do in more complex examples. For example, in Gauss's Disq. Arith. the proof of associativity of composition of binary quadratic forms comprises many pages of unilluminating abstruse calculations, whereas nowadays this can be done by simply by transporting the class group structure from ideals to primitive binary quadratic forms.

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