Proof of Theorem KNSI

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

We must first insure that if we choose an input for $\vectrepname{B}$ from $\krn{T}$ that then the output will be an element of $\nsp{\matrixrep{T}{B}{C}}$. So suppose that $\vect{u}\in\krn{T}$. Then \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}{\zerovector}&&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &=\zerovector&&\knowl{./knowls/theorem.LTTZZ.knowl}{\text{Theorem LTTZZ}}\\ \end{align*}

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

The restriction in the size of the domain and codomain $\vectrepname{B}$ will not affect the fact that $\vectrepname{B}$ is a linear transformation (Theorem VRLT), nor will it affect the fact that $\vectrepname{B}$ is injective (Theorem VRI). Something must be done though to verify that $\vectrepname{B}$ is surjective. To this end, appeal to the definition of surjective (Definition SLT), and suppose that we have an element of the codomain, $\vect{x}\in\nsp{\matrixrep{T}{B}{C}}\subseteq\complex{n}$ and we wish to find an element of the domain with $\vect{x}$ as its image. We now show that the desired element of the domain is $\vect{u}=\vectrepinv{B}{\vect{x}}$. First, verify that $\vect{u}\in\krn{T}$, \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}{\zerovector_{\complex{n}}}&&\knowl{./knowls/definition.KLT.knowl}{\text{Definition KLT}}\\ &=\zerovector_{V}&&\knowl{./knowls/theorem.LTTZZ.knowl}{\text{Theorem LTTZZ}}\\ \end{align*} Second, verify that the proposed isomorphism, $\vectrepname{B}$, takes $\vect{u}$ to $\vect{x}$, \begin{align*} \vectrep{B}{\vect{u}}&=\vectrep{B}{\vectrepinv{B}{\vect{x}}}&&\text{Substitution}\\ &=\lt{I_{\complex{n}}}{\vect{x}}&&\knowl{./knowls/definition.IVLT.knowl}{\text{Definition IVLT}}\\ &=\vect{x}&&\knowl{./knowls/definition.IDLT.knowl}{\text{Definition IDLT}} \end{align*}

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