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Show that $\forall a\in\mathbb{H}, \ \exists b \in\mathbb{H}: ab =ba = 1.$

I am pretty sure I can easily google the multiplicative inverse in $\mathbb{H}$, but can you give me a hint on how to determine the inverse von an arbitrary $a \in\mathbb{h}$ myself? This task comes directly after showing that the conjugation is a ring antihomomorphism. ($a\ \bar+ \ b\ = \bar a+\bar b, a\ \bar *\ b = \bar b*\bar a$ and $ 1 = \bar1$)

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    $\begingroup$ Need $a\neq 0$. $\endgroup$
    – user99914
    Commented Dec 9, 2013 at 2:15
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    $\begingroup$ Do you know how to prove the corresponding statement for $\mathbb{C}$? $\endgroup$ Commented Dec 9, 2013 at 2:15
  • $\begingroup$ @JasonDeVito $z^{-1} = \frac{\bar z}{\bar zz}$? I tried that one an hour ago as well, but I got something like $\bar zz = (... + 0i + 0j + 0k)$ so in $\frac{\bar z}{\bar zz}$ I end up dividing by zero. $\endgroup$
    – Nhat
    Commented Dec 9, 2013 at 2:48
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    $\begingroup$ @kitkat4.4: Check the $...$ portion again - you only end up dividing by $0$ when $z$ is $0$. $\endgroup$ Commented Dec 9, 2013 at 4:04

3 Answers 3

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This is a somewhat nonstandard approach, but I think it illustrates a nice model for $\mathbb H$.

Note that $\mathbb H$ is isomorphic to the algebra of matrices over $\mathbb C$ of the form $$\begin{pmatrix} a+bi & c+di\\ -c+di & a-bi \end{pmatrix}$$ which is easy to verify: you just need to check that it contains a copy of $\mathbb R$ (namely the matrices with $b=c=d=0$) and that it contains $i,j,k$ such that $i^2=j^2=k^2=ijk=-1$, which come from setting everything but $b,c$ and $d$ respectively equal to $0$.

Now the inverse is just the inverse of a $2\times 2$ matrix, which is easy to compute and can easily be shown to have the same form.

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  • $\begingroup$ This is an interesting halfway point between the other two solutions :) Let's say that one found the inverse of this matrix using the adjugate matrix. If the matrix above is $q$, then the quaternion conjugate $\bar q$ is the adjugate of the above matrix, and the determinant is $q\bar q$. So the adjugate divided by the determinant gives you the inverse, and this is $\bar q /q\bar q$. (The model of $\Bbb H$ I used was... $\Bbb H$ :) ) $\endgroup$
    – rschwieb
    Commented Dec 9, 2013 at 19:42
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Show that $\forall a\in\mathbb{H}, \ \exists b \in\mathbb{H}: ab =ba = 1.$

This task comes directly after showing that the conjugation is a ring antihomomorphism.

Inverses are easily found inside the quaternions with conjugation, multiplication and division.

Really, the only computational fact you need is that $a\overline{a}\in \Bbb R$. By looking at what conjugation does to quaternions, you can easily see that $\overline{q}=q\iff q\in \Bbb R$.

But then since conjugation is an antihomomorphism, $\overline{q\overline{q}}=\overline{\bar q}\bar q=q\bar q$, so $q\bar q$ is real.

But then look: $q(\frac{\bar q}{q\bar q})=1$ finds a right inverse for $q$. Since this is true for all nonzero quaternions, $\frac{\bar q}{q\bar q}$ has a right inverse also, which is necessarily $q$. Thus $\frac{\bar q}{q\bar q}=q^{-1}$.

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  • $\begingroup$ I marked your solution "the answer" because when my tutor was dropping hints for some solutions, he said word-by-word what you posted in your first two sentences. - Obviously, this was the hint I was looking for today, but all the other solutions are probably just as good as yours. (Although the one with the four equations is cringeworthy.) $\endgroup$
    – Nhat
    Commented Dec 9, 2013 at 23:33
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You should just solve for ($(x + y i + z j + w k)(a+b i + c j + d k) = 1.$ Notice that you will get a system of linear equations in $x, y, z, w.$

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  • $\begingroup$ And the determinant of the system is, guess what, $a^2+b^2+c^2+d^2$. $\endgroup$
    – egreg
    Commented Dec 9, 2013 at 14:49
  • $\begingroup$ Dear Igor: I mean no disrespect by this comment, but one could say that this is one of the the hardest correct solutions possible. It is possible to solve in a few lines without resulting to a large system of linear equations. Regards $\endgroup$
    – rschwieb
    Commented Dec 9, 2013 at 14:49
  • $\begingroup$ @rschwieb I disagree. I agree that it is possible to VERIFY the solution in a couple of lines, but to arrive at it, I don't know of an easier way. $\endgroup$
    – Igor Rivin
    Commented Dec 9, 2013 at 18:54
  • $\begingroup$ Dear @IgorRivin : Yes, this is a practical solution! It's certainly worthy of the OP's acceptance. But really I just want to draw your attention to the trick that Jason Devito and I are pointing at so that in the future you can recommend something better than brute force :) Regards $\endgroup$
    – rschwieb
    Commented Dec 9, 2013 at 19:29
  • $\begingroup$ It's really interesting though: we essentially have three different solutions approaching the problem the same way, but one approaches $\Bbb H$ as a matrix ring over $\Bbb R$, Alex's approaches it as a matrix ring with entries in $\Bbb C$, and mine approaches it directly as $\Bbb H$... very pretty array of solutions! $\endgroup$
    – rschwieb
    Commented Dec 9, 2013 at 19:43

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