Question

Let 1, i, j, k be the following matrices with complex entries:

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

(a) Prove that

$i^2=j^2=k^2=-1$ $ij=-ji=k$

$jk=-kj=i$ $ki=-ik=j$

Short answer

It is proved that $i^2=j^2=k^2=-1$, $ij=-ji=k$, $jk=-kj=i$ and$ki=-ik=j$.

Step-by-step solution

Obtaining i2=j2=k2=-1 

Consider the given matrices, $1=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right),i=\left(\begin{array}{cc}i& 0\\ 0& -i\end{array}\right),j=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right),k=\left(\begin{array}{cc}0& i\\ i& 0\end{array}\right)$

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

Also,

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

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

Obtain ij=-ji=k and jk=-kj=i    

Therefore,

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

Similarly,

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

And

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

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

Obtain ki=-ik=j

Thus,

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

Similarly,

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