Proof of Theorem MIMI

As with the proof of Theorem SS, we examine if $A$ is a suitable inverse for $\inverse{A}$ (by definition, the opposite is true). \begin{align*} A\inverse{A}&=I_n&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}} \end{align*} and \begin{align*} \inverse{A}A&=I_n&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}} \end{align*}

The matrix $A$ has met all the requirements to be the inverse of $\inverse{A}$, and so is invertible and we can write $A=\inverse{(\inverse{A})}$.