Let H be the skew field of quaternions.
Find all quaternions x satisfying $(i + j)x(i + k) = 2$
I'm having trouble figuring out what to do with this question. I know the "i j k i j k" formula for determining the cross-product of vectors. $(ij = k, i^{2} = -1, etc)$
Am I supposed to play around with the multiplicativity until I end up with something like $((-1) + (-1))(-1) = 2?$
Thanks in advance for the help.
On
There is a general way to solve an equation of the type $$axb+cxd=e$$ in the skew field of quaternions. The trick is to note that, given a quaternion $z=z_0+z_1\mathbf{i}+z_2\mathbf{j}+z_3\mathbf{k}$ and its conjugate $\bar z=z_0-z_1\mathbf{i}-z_2\mathbf{j}-z_3\mathbf{k}$, then $z \bar z=|z|^2$ and $z+\bar z=2z_0$ are real numbers, then commute with all quaternions. So, multiplying left by $\bar c$ and right by $\bar b$ we have $$ axb+cxd=e \iff |b|^2\bar c a x+x d \bar b |c|^2=\bar c e \bar b $$ so the equation reduces to the form $$ \alpha x+ x \beta = \gamma $$ and for the solution of this equation you can use the same trick and the fact that $ \alpha(\alpha x+ x \beta) = \alpha \gamma$ obtaining the formula in Solution of $ax+xb=c$ in a division ring
As you've mentioned, the quaternions form a skew field. Thus the element $(i+j)$ has an inverse $a$ (so that $a(i+j)=1$), and there exists a $b$ such that $(i+k)b=1$. Then $$a(i+j)x(i+k)b=x=2ab$$ Thus there is exactly one solution. Note that in writing this equation I used the fact that $2$ commutes with every quaternion, as does any real number.
The goal now is to find $a$ and $b$. I'd start by trying to solve $$(c+di+ej+fk)(i+j)=1$$ where $c,d,e,f$ are real by expanding the product and equating the real parts as well as the corresponding coefficients of $i,j,k$. There should be only one solution. Then do the same for $i+k$, then multiply the results to get your answer.