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:

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

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

role="math" localid="1659340998671" $\begin{array}{rcl}\mathit{a}\mathbf{1}\mathbf{+}\mathit{b}\mathit{i}\mathbf{+}\mathit{c}\mathit{j}\mathbf{+}\mathit{d}\mathit{k}& \mathbf{=}& \mathit{a}\mathbf{\left(}\begin{array}{cc}\mathbf{1}& \mathbf{0}\\ \mathbf{0}& \mathbf{1}\end{array}\mathbf{\right)}\mathbf{+}\mathit{b}\mathbf{\left(}\begin{array}{cc}\mathbf{i}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{+}\mathit{c}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{1}\\ \mathbf{-}\mathbf{1}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{+}\mathit{d}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{i}\\ \mathbf{i}& \mathbf{0}\end{array}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{\left(}\begin{array}{cc}\mathbf{a}& \mathbf{0}\\ \mathbf{0}& \mathbf{a}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{bi}& \mathbf{0}\\ \mathbf{0}& \mathbf{-}\mathit{b}\mathbf{i}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& c\\ \mathbf{-}\mathit{c}& \mathbf{0}\end{array}\mathbf{\right)}\mathbf{+}\mathbf{\left(}\begin{array}{cc}\mathbf{0}& \mathbf{di}\\ \mathbf{di}& \mathbf{0}\end{array}\mathbf{\right)}\\ & \mathbf{=}& \left(\begin{array}{cc}a+bi& c+di\\ -c+di& a-bi\end{array}\right)\end{array}$

Where a,b,c are d real numbers.

1. Prove that
   role="math" localid="1659341258137" $$\begin{gathered} {\mathit{i}}^{\mathbf{2}}\mathbf{=}{\mathit{j}}^{\mathbf{2}}\mathbf{=}{\mathit{k}}^{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1} \\ \mathit{j}\mathit{k}\mathbf{=}\mathbf{-}\mathit{k}\mathit{j}\mathbf{=}\mathit{i} \end{gathered}$$ $$\begin{gathered} \mathit{i}\mathit{j}\mathbf{=}\mathbf{-}\mathit{j}\mathit{i}\mathbf{=}\mathit{k} \\ \mathit{k}\mathit{i}\mathbf{=}\mathbf{-}\mathit{i}\mathit{k}\mathbf{=}\mathit{j} \end{gathered}$$

Short answer

It is proved that i² = j² = k² = -1 , ij = -ji = k , jk = -kj = i , and ki = j .

Step-by-step solution

Prove that i2 = j2 = k2 = -1 

Prove the result as:

$$\begin{gathered} i^2={\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)}^2 \\ =\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right) \\ =\left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right) \\ =-\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right) \\ =-1 \end{gathered}$$

$\begin{array}{rcl}{j}^2& =& {\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)}^2\\ & =& \left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ & =& -\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ & =& -1\end{array}$

$\begin{array}{rcl}{k}^2& =& {\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)}^2\\ & =& \left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right)\\ & =& -\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ & =& -1\end{array}$

Thus, it is proved that i² = j² = k² = -1 .

Prove that ij = -ji = k 

First, prove that ij = -ji = k as:

$\begin{array}{rcl}ij& =& \left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\\ & =& k\end{array}$

$\begin{array}{rcl}-ji& =& -\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\\ & =& -\left(\begin{array}{cc}0& -i\\ -i& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\\ & =& k\end{array}$

Thus, ij = -ji = k.

Prove that jk = -kj = i 

First, prove that jk = -kj = i as:

$\begin{array}{rcl}jk& =& \left(\begin{array}{cc}0& 1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\\ & =& i\end{array}$

$\begin{array}{rcl}kj& =& \left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}-i& 0\\ 0& i\end{array}\right)\\ & =& -\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\\ & =& -i\end{array}$

Thus, jk = -kj = i.

Prove that ki = -ik = j 

First, prove that ki = -ik = j as:

$\begin{array}{rcl}ki& =& \left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\\ & =& \left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ & =& j\\ ik& =& \left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right)\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)\\ & =& \left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\\ & =& -\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)\\ & =& -j\end{array}$

Thus, ki = -ik = j.

Hence it is proved that ij = -ji = k , jk = - kj = i, and ki = -ik = j .