Benedict Irwin edited Complex.tex  over 9 years ago

Commit id: 1d12e7004664211c6f9dc3e9a19696180fd9e351

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\begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}\#\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}  \end{equation}  However it is not possible to define a right identity! Using the same matrix above will keep the elements the same bu swap those in the right hand column. Most likely a strange artifact of the way we came to the transform. We can define a left inverse, for some $A\#B=C$ such that $(A^{-1}\#A)\#B=A^{-1}\#C=B$. For such a system we solve for the $a,b,c,d$ such that \begin{equation}  \begin{bmatrix} e & 0 & 0 &-h \\ f & 0 & 0 & g \\ 0 & -f & g & 0 \\ 0 & e & h & 0 \end{bmatrix}\begin{bmatrix}a \\ b \\ c \\ d \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}  \end{equation}  Which results in \begin{equation}  A^{-1}=\frac{1}{eg+fh}\begin{bmatrix}g&h&0&0\\0&0&-h&g\\0&0&e&f\\-f&e&0&0\end{bmatrix}  \end{equation}\begin{bmatrix}1 \\ 0 \\ 1 \\ 0 end{bmatrix}