Proof of Theorem MIT

As with the proof of Theorem SS, we see if $\transpose{(\inverse{A})}$ is a suitable inverse for $\transpose{A}$. Apply Theorem MMT to see that \begin{align*} \transpose{(\inverse{A})}\transpose{A} &=\transpose{(A\inverse{A})}&&\knowl{./knowls/theorem.MMT.knowl}{\text{Theorem MMT}}\\ &=\transpose{I_n}&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=I_n&&\knowl{./knowls/definition.SYM.knowl}{\text{Definition SYM}} \end{align*} and \begin{align*} \transpose{A}\transpose{(\inverse{A})} &=\transpose{(\inverse{A}A)}&&\knowl{./knowls/theorem.MMT.knowl}{\text{Theorem MMT}}\\ &=\transpose{I_n}&&\knowl{./knowls/definition.MI.knowl}{\text{Definition MI}}\\ &=I_n&&\knowl{./knowls/definition.SYM.knowl}{\text{Definition SYM}} \end{align*}

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