Proof of Theorem RCSI

To establish that two vector spaces are isomorphic, we must find an isomorphism between them, an invertible linear transformation (Definition IVS). The range of the linear transformation $T$, $\rng{T}$, is a subspace of $V$, while the column space of the matrix representation, $\csp{\matrixrep{T}{B}{C}}$ is a subspace of $\complex{m}$. The function $\vectrepname{C}$ is defined as a function from $V$ to $\complex{m}$, but we can just as well employ the definition of $\vectrepname{C}$ as a function from $\rng{T}$ to $\csp{\matrixrep{T}{B}{C}}$.

We must first insure that if we choose an input for $\vectrepname{C}$ from $\rng{T}$ that then the output will be an element of $\csp{\matrixrep{T}{B}{C}}$. So suppose that $\vect{v}\in\rng{T}$. Then there is a vector $\vect{u}\in U$, such that $\lt{T}{\vect{u}}=\vect{v}$. Consider \begin{align*} \matrixrep{T}{B}{C}\vectrep{B}{\vect{u}} &=\vectrep{C}{\lt{T}{\vect{u}}}&&\knowl{./knowls/theorem.FTMR.knowl}{\text{Theorem FTMR}}\\ &=\vectrep{C}{\vect{v}}&&\knowl{./knowls/definition.RLT.knowl}{\text{Definition RLT}} \end{align*}

This says that $\vectrep{C}{\vect{v}}\in\csp{\matrixrep{T}{B}{C}}$, as desired.

The restriction in the size of the domain and codomain will not affect the fact that $\vectrepname{C}$ is a linear transformation (Theorem VRLT), nor will it affect the fact that $\vectrepname{C}$ is injective (Theorem VRI). Something must be done though to verify that $\vectrepname{C}$ is surjective. This all gets a bit confusing, since the domain of our isomorphism is the range of the linear transformation, so think about your objects as you go. To establish that $\vectrepname{C}$ is surjective, appeal to the definition of a surjective linear transformation (Definition SLT), and suppose that we have an element of the codomain, $\vect{y}\in\csp{\matrixrep{T}{B}{C}}\subseteq\complex{m}$ and we wish to find an element of the domain with $\vect{y}$ as its image. Since $\vect{y}\in\csp{\matrixrep{T}{B}{C}}$, there exists a vector, $\vect{x}\in\complex{n}$ with $\matrixrep{T}{B}{C}\vect{x}=\vect{y}$.

We now show that the desired element of the domain is $\vect{v}=\vectrepinv{C}{\vect{y}}$. First, verify that $\vect{v}\in\rng{T}$ by applying $T$ to $\vect{u}=\vectrepinv{B}{\vect{x}}$, \begin{align*} \lt{T}{\vect{u}}&=\lt{T}{\vectrepinv{B}{\vect{x}}}\\ &=\vectrepinv{C}{\matrixrep{T}{B}{C}\left(\vectrep{B}{\vectrepinv{B}{\vect{x}}}\right)}&&\knowl{./knowls/theorem.FTMR.knowl}{\text{Theorem FTMR}}\\ &=\vectrepinv{C}{\matrixrep{T}{B}{C}\left(\lt{I_{\complex{n}}}{\vect{x}}\right)}&&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\vectrepinv{C}{\matrixrep{T}{B}{C}\vect{x}}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}}\\ &=\vectrepinv{C}{\vect{y}}&&\knowl{./knowls/definition.CSM.knowl}{\text{Definition CSM}}\\ &=\vect{v}&&\text{Substitution}\\ \end{align*} Second, verify that the proposed isomorphism, $\vectrepname{C}$, takes $\vect{v}$ to $\vect{y}$, \begin{align*} \vectrep{C}{\vect{v}}&=\vectrep{C}{\vectrepinv{C}{\vect{y}}}&&\text{Substitution}\\ &=\lt{I_{\complex{m}}}{\vect{y}}&&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\vect{y}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}} \end{align*}

With $\vectrepname{C}$ demonstrated to be an injective and surjective linear transformation from $\rng{T}$ to $\csp{\matrixrep{T}{B}{C}}$, Theorem ILTIS tells us $\vectrepname{C}$ is invertible, and so by Definition IVS, we say $\rng{T}$ and $\csp{\matrixrep{T}{B}{C}}$ are isomorphic.