What is the inverse element of $u=i+j+k$ with respect to the multiplication?

Question

What is the inverse element of $u=i+j+k$ with respect to the multiplication?

When I have $z = x + iy$ inverse is $\frac{x}{x^2+y^2}-\frac{iy}{x^2+y^2}$ but what happens in my case?

Answer 1

Try $\frac{-i-j-k}{3}$. This is a natural extension of the complex conjugate formulation to the context of having three imaginary units.

I think Don Antonio's answer gives a good indication of how to find the inverse. To give another example of this sort of thinking: in the hyperbolic numbers $z = x+jy$ where $j^2=1$ we find the multiplicative inverse of $a+bj$ by solving $$ (a+bj)(x+jy) = ax+by+j(ay+bx) = 1 = 1+j(0)$$ for $x+jy$. In particular, we must solve real equations $ax+by=1$ and $bx+ay = 0$: $$ \left[ \begin{array}{cc} a & b \\ b & a \end{array} \right] \left[ \begin{array}{c}x \\ y \end{array} \right] = \left[ \begin{array}{c}1 \\ 0 \end{array} \right] \ \ \Rightarrow \ \ \left[ \begin{array}{c}x \\ y \end{array} \right] = \frac{1}{a^2-b^2}\left[ \begin{array}{cc} a & -b \\ -b & a \end{array} \right]\left[ \begin{array}{c}1 \\ 0 \end{array} \right] = \frac{1}{a^2-b^2}\left[ \begin{array}{c} a \\ -b \end{array} \right] $$ That is, $$(a+bj)^{-1} = \frac{a-bj}{a^2-b^2}$$ Notice, the matrix $$ \left[ \begin{array}{cc} a & b \\ b & a \end{array} \right] $$ naturally corresponds to $a+bj$ via the map $\Psi(a+bj) = \left[ \begin{array}{cc} a & b \\ b & a \end{array} \right]$ and you can easily show $\Psi(zw) = \Psi(z) \Psi(w)$, $\Psi(z+w) = \Psi(z)+ \Psi(w)$ and $\Psi(1) = \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right]$. This is the regular representation of hyperbolic numbers.

Generally, given a real associative algebra, a nice way to find the inverse is given by the regular representation. Simply find the corresponding matrix to your algebra element then use standard linear algebra to find the inverse of the matrix. Furthermore, this also reveals the structure of zero divisors in the algebra. With the hyperbolic numbers, the zero divisors are those $a+bj$ for which $a^2=b^2$. In contrast, the quaternions are a skew field where every nonzero element has a multiplicative inverse.

Answer 2

Do the maths: if $\;a+bi+cj+dk\;$ is the inverse, then it must be that

$$1=(i+j+k)(a+bi+cj+dk)=ai-b+ck-dj+aj-bk-c+di+ak+bj-ci-d$$

$$=(-b-c-d)+(a-c+d)i+(a+b-d)j+(a-b+c)k\implies\text{ now solve the corresponding linear system}$$

Answer 3

For any nonzero quaternion $q$, obviously we have* $$q\frac{\bar{q}}{q\bar q}=1$$

Said another way: $q^{-1}=\frac{\bar q}{q\bar q}$

In your case, $\bar q=-i-k-j$ and $q\bar q=3$. ($q\bar q$ is always equal to the squared Euclidean length of the coefficient vector.)

*_{There is a tiny bit of voodoo going on here with writing fractions over a skew field like $\mathbb H$. But since the denominator is a real number, the expression is 'saved.' If you really like, you can rewrite this to be $q(\bar q(q\bar q)^{-1})=1$ which is technically best but less readable.}

Answer 4

In tessarines, it is

$$\frac1{i+j+ij}=-\left(\frac{2}{5}+\frac{i}{5}\right)+\left(\frac{3}{5}-\frac{i}{5}\right)j$$