Proof of Theorem IMR

(⇒) Suppose $T$ is invertible, so the inverse linear transformation $\ltdefn{\ltinverse{T}}{V}{U}$ exists (Definition IVLT). Both linear transformations have matrix representations relative to the bases of $U$ and $V$, namely $\matrixrep{T}{B}{C}$ and $\matrixrep{\ltinverse{T}}{C}{B}$ (Definition MR).

Then \begin{align*} \matrixrep{\ltinverse{T}}{C}{B}\matrixrep{T}{B}{C} &=\matrixrep{\compose{\ltinverse{T}}{T}}{B}{B} &&\knowl{./knowls/theorem.MRCLT.knowl}{\text{Theorem MRCLT}}\\ &=\matrixrep{I_U}{B}{B} &&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\left[ \left.\vectrep{B}{\lt{I_U}{\vect{u}_1}}\right| \left.\vectrep{B}{\lt{I_U}{\vect{u}_2}}\right| \ldots \left|\vectrep{B}{\lt{I_U}{\vect{u}_n}}\right. \right] &&\knowl{./knowls/definition.MR.knowl}{\text{Definition MR}}\\ &=\left[ \left.\vectrep{B}{\vect{u}_1}\right| \left.\vectrep{B}{\vect{u}_2}\right| \left.\vectrep{B}{\vect{u}_3}\right| \ldots \left|\vectrep{B}{\vect{u}_n}\right. \right] &&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\matrixcolumns{e}{n} &&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}\\ &=I_n &&\knowl{./knowls/definition.IM.knowl}{\text{Definition IM}} \end{align*}

And \begin{align*} \matrixrep{T}{B}{C}\matrixrep{\ltinverse{T}}{C}{B}&= \matrixrep{\compose{T}{\ltinverse{T}}}{C}{C} &&\knowl{./knowls/theorem.MRCLT.knowl}{\text{Theorem MRCLT}}\\ &=\matrixrep{I_V}{C}{C} &&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\left[ \left.\vectrep{C}{\lt{I_V}{\vect{v}_1}}\right| \left.\vectrep{C}{\lt{I_V}{\vect{v}_2}}\right| \ldots \left|\vectrep{C}{\lt{I_V}{\vect{v}_n}}\right. \right] &&\knowl{./knowls/definition.MR.knowl}{\text{Definition MR}}\\ &=\left[ \left.\vectrep{C}{\vect{v}_1}\right| \left.\vectrep{C}{\vect{v}_2}\right| \left.\vectrep{C}{\vect{v}_3}\right| \ldots \left|\vectrep{C}{\vect{v}_n}\right. \right]&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\matrixcolumns{e}{n} &&\knowl{./knowls/definition.VR.knowl}{\text{Definition VR}}\\ &=I_n &&\knowl{./knowls/definition.IM.knowl}{\text{Definition IM}} \end{align*}

These two equations show that $\matrixrep{T}{B}{C}$ and $\matrixrep{\ltinverse{T}}{C}{B}$ are inverse matrices (Definition MI) and establish that when $T$ is invertible, then $\matrixrep{\ltinverse{T}}{C}{B}=\inverse{\left(\matrixrep{T}{B}{C}\right)}$.

(⇐) Suppose now that $\matrixrep{T}{B}{C}$ is an invertible matrix and hence nonsingular (Theorem NI). We compute the nullity of $T$, \begin{align*} \nullity{T} &=\dimension{\krn{T}} &&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &=\dimension{\nsp{\matrixrep{T}{B}{C}}} &&\knowl{./knowls/theorem.KNSI.knowl}{\text{Theorem KNSI}}\\ &=\nullity{\matrixrep{T}{B}{C}} &&\knowl{./knowls/definition.NOM.knowl}{\text{Definition NOM}}\\ &=0 &&\knowl{./knowls/theorem.RNNM.knowl}{\text{Theorem RNNM}} \end{align*}

So the kernel of $T$ is trivial, and by Theorem KILT, $T$ is injective.

We now compute the rank of $T$, \begin{align*} \rank{T} &=\dimension{\rng{T}} &&\knowl{./knowls/definition.RLT.knowl}{\text{Definition RLT}}\\ &=\dimension{\csp{\matrixrep{T}{B}{C}}} &&\knowl{./knowls/theorem.RCSI.knowl}{\text{Theorem RCSI}}\\ &=\rank{\matrixrep{T}{B}{C}} &&\knowl{./knowls/definition.ROM.knowl}{\text{Definition ROM}}\\ &=\dimension{V} &&\knowl{./knowls/theorem.RNNM.knowl}{\text{Theorem RNNM}} \end{align*}

Since the dimension of the range of $T$ equals the dimension of the codomain $V$, by Theorem EDYES, $\rng{T}=V$. Which says that $T$ is surjective by Theorem RSLT.

Because $T$ is both injective and surjective, by Theorem ILTIS, $T$ is invertible.