Question

In the ring M(C), let

$\mathbf{1}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{i}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{i}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathit{j}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{}\mathbf{}\mathit{k}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{i}\\ \mathbf{i}& \mathbf{0}\end{array}\mathbf{\right)}$

The Product of a real number and a matrix is the matrix given by this rule:

$\mathit{r}\mathbf{\left(}\begin{array}{cc}\mathbf{t}& \mathbf{u}\\ \mathbf{v}& \mathbf{w}\end{array}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\begin{array}{cc}\mathbf{rt}& \mathbf{ru}\\ \mathbf{rv}& \mathbf{rw}\end{array}\mathbf{\right)}$

The set H of real quaternions consists of all matrices of the form

$\begin{array}{rcl}\mathit{a}\mathbf{1}\mathbf{+}\mathit{b}\mathit{i}\mathbf{+}\mathit{c}\mathit{j}\mathbf{+}\mathit{d}\mathit{k}& \mathbf{=}& \mathit{a}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{+}\mathit{b}\mathbf{\left(}\begin{array}{cc}\mathbf{i}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{+}\mathit{c}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{+}\mathit{d}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{i}\\ \mathbf{i}& \mathbf{0}\end{array}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{\left(}\begin{array}{cc}\mathbf{a}& \mathbf{0}\\ \mathbf{0}& \mathbf{a}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{bi}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathit{b}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& c\\ \mathbf{-}\mathit{c}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{di}\\ \mathbf{di}& \mathbf{0}\end{array}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{\left(}\begin{array}{cc}\mathbf{a}\mathbf{+}\mathbf{bi}& \mathbf{c}\mathbf{+}\mathbf{di}\\ \mathbf{-}\mathbf{c}\mathbf{+}\mathbf{di}& \mathbf{a}\mathbf{-}\mathbf{bi}\end{array}\mathbf{\right)}\end{array}$

Where a,b,c are d real numbers.

(d) Show that the equation x² = -1 has infinitely many solutions in H. [Hint: Consider quaternions of the form 01 + bi + cj - dk where b² + c² + d² + = 1.]

Short answer

It is proved that x² = -1 has infinitely many solutions in H.

Step-by-step solution

Define

Consider to be some real quaternion of the form $A=01+bi+cj-dk$where $b^2+c^2+d^2=1$.

Prove that x2 = -1 has infinitely many solutions in H 

Solve for A² as:

$\begin{array}{rcl}{A}^2& =& \left(01+bi+cj-dk\right)\left(01+bi+cj-dk\right)\\ & =& -b^{2}1+bck+bdj-bck-c^{2}1-cdi-bdj+cdi-d^{2}1\\ & =& -b^{2}1-c^{2}1-d^{2}1\\ & =& -\left(b^2+c^2+d^2\right)1\\ & =& -1\end{array}$

This implies that there are infinitely many real numbers b, c, and d such that b² + c² + d² = 1 , which implies that there are infinitely many real quaternions A such that A² = -1.