四元数の行列表現での逆元

四元数の行列表現での逆元

$$ \begin{eqnarray} \mathbf{q}=w+xi+yj+zk &\leftrightarrow& \href{https://shikitenkai.blogspot.com/2020/06/blog-post_99.html}{ \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} } \end{eqnarray} $$ $$ \begin{eqnarray} |\mathbf{q}|^2=|w+xi+yj+zk|^2 &\leftrightarrow& \begin{vmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{vmatrix} \\&=&(w+xi)(w-xi)-(y+zi)(-(y-zi)) \\&=&(w+xi)(w-xi)+(y+zi)(y-zi) \\&=&w^2+x^2+y^2+z^2 \end{eqnarray} $$ $$ \begin{eqnarray} \bar{\mathbf{q}}=w-xi-yj-zk &\leftrightarrow& \begin{bmatrix} w+(-x)i&(-y)+(-z)i\\ -((-y)-(-z)i)&w-(-x)i\\ \end{bmatrix} \\&=& \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{q}\bar{\mathbf{q}} &=&(w+xi+yj+zk)(w-xi-yj-zk) \\&\leftrightarrow& \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \\&=& \begin{bmatrix} (w+xi)(w-xi)+(y+zi)(-(-y+zi)) &(w+xi)(-y-zi)+(y+zi)(w+xi)\\ (-(y-zi))(w-xi)+(w-xi)(-(-y+zi)) &(-(y-zi))(-y-zi)+(w-xi)(w+xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} (w+xi)(w-xi)+(y+zi)(y-zi) &(w+xi)(-y-zi)+(y+zi)(w+xi)\\ (-y+zi)(w-xi)+(w-xi)(y-zi) &(-y+zi)(-y-zi)+(w-xi)(w+xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2-wxi+wxi+x^2 +y^2-yzi+yzi+z^2 &-wy-wzi-xyi+xz +wy+xyi+wzi-xz\\ -wy+xyi+wzi+xz +wy-wzi-xyi-xz &y^2+yzi-yzi+z^2 +w^2+wxi-wxi+x^2\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2+x^2+y^2+z^2 &0\\ 0 &w^2+x^2+y^2+z^2\\ \end{bmatrix} \end{eqnarray} $$ $$ \begin{eqnarray} \bar{\mathbf{q}}\mathbf{q} &=&(w-xi-yj-zk)(w+xi+yj+zk) \\&\leftrightarrow& \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \begin{bmatrix} w+xi&y+zi\\ -(y-zi)&w-xi\\ \end{bmatrix} \\&=& \begin{bmatrix} (w-xi) (w+xi) +(-y-zi) (-(y-zi)) & (w-xi) (y+zi) +(-y-zi) (w-xi)\\ (-(-y+zi)) (w+xi) +(w+xi) (-(y-zi)) & (-(-y+zi)) (y+zi) +(w+xi) (w-xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} (w-xi) (w+xi) +(-y-zi) (-y+zi) & (w-xi) (y+zi) +(-y-zi) (w-xi)\\ (y-zi) (w+xi) +(w+xi) (-y+zi) & (y-zi) (y+zi) +(w+xi) (w-xi)\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2 +wxi -wxi +x^2 +y^2 +yzi -yzi +z^2 &wy +wzi -xyi +xz -wy +xyi -wzi -xz\\ wy +xyi -wzi +xz -wy +wzi -xyi -xz &y^2 +yzi -yzi +z^2 +w^2 -wxi +wxi +x^2\\ \end{bmatrix} \\&=& \begin{bmatrix} w^2+x^2+y^2+z^2 &0\\ 0 &w^2+x^2+y^2+z^2\\ \end{bmatrix} \;\cdots\;行列の積は一般に非可換だが，この形の行列では可換である \end{eqnarray} $$ $$ \begin{eqnarray} \frac{\mathbf{q}\bar{\mathbf{q}}}{|\mathbf{q}|^2} &=&\frac{\bar{\mathbf{q}}\mathbf{q}}{|\mathbf{q}|^2} =\mathbf{q}\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2} =\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2}\mathbf{q} =\frac{1}{|\mathbf{q}|^2}\mathbf{q}\bar{\mathbf{q}} =\frac{1}{|\mathbf{q}|^2}\bar{\mathbf{q}}\mathbf{q} \\&\leftrightarrow& \frac{1}{w^2+x^2+y^2+z^2} \begin{bmatrix} w^2+x^2+y^2+z^2 &0\\ 0 &w^2+x^2+y^2+z^2\\ \end{bmatrix} \\&=& \begin{bmatrix} 1&0\\ 0&1\\ \end{bmatrix} \\&=&\mathbf{E} \end{eqnarray} $$ $$ \begin{eqnarray} \mathbf{q}^{-1} &=&\frac{\bar{\mathbf{q}}}{|\mathbf{q}|^2} \\&\leftrightarrow& \frac{1}{w^2+x^2+y^2+z^2} \begin{bmatrix} w-xi&-y-zi\\ -(-y+zi)&w+xi\\ \end{bmatrix} \end{eqnarray} $$