% This is lecture no. 10 of MH1201/: Linear Algebra II.

\chapter{Invertibility \& Isomorphisms}%
\label{cha:invertibility_&_isomorphisms}

\begin{definition}
	Let $\mathsf{V}$ and $\mathsf{W}$ be vector spaces, and let $\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$.
	Then a function $\mathsf{U}:\mathsf{W}\mapsto\mathsf{V}$ is said to be an \textbf{inverse} of $\mathsf{T}$ if:
	\[
		\mathsf{T}\circ\mathsf{U}=I_{\mathsf{W}}
		\quad
		\text{and}
		\quad
		\mathsf{U}\circ\mathsf{T}=I_{\mathsf{V}}
	.\]
	If $\mathsf{T}$ has an inverse, then $\mathsf{T}$ is said to be \textbf{invertible}.
\end{definition}

We know from \texttt{MH1300} that if $\mathsf{T}$ is invertible, then its inverse is unique and is denoted by $\mathsf{T}^{-1}$.
We also know that it implies that $\mathsf{T}$ is one-to-one and onto.

\begin{thrm}
	Let $\mathsf{V}$ and $\mathsf{W}$ be finite-dimensional vector spaces where $\mdim(\mathsf{V})=\mdim(\mathsf{W})$, and let $\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$.
	Then:
	\[
	\mathsf{T}\ \text{invertible}\ \iff\mdim(R(\mathsf{T}))=\mdim(\mathsf{V})
	.\]
	Since invertibility implies one-to-one and onto.
\end{thrm}
The following result shows that the inverse of a linear transformation preserves linearity.
\begin{thrm}
	Given an invertible $\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$ we conclude that $\mathsf{T}^{-1}:\mathsf{W}\mapsto\mathsf{V}$ is also linear, that is, $\mathsf{T}^{-1}\in\mathcal{L}(\mathsf{W},\mathsf{V})$.
\end{thrm}
\begin{proof}
	Let $w_1,w_2\in\mathsf{W}$ and $\lambda\in F$.
	Since $\mathsf{T}$ is onto and one-to-one, there exists unique vectors  $v_1,v_2\in\mathsf{V}$ such that:
	\[
	\mathsf{T}(v_1)=w_1
	\quad
	\text{and}
	\quad
	\mathsf{T}(v_2)=w_2
	.\]
	Thus:
	\[
	v_1=\mathsf{T}^{-1}(w_1)
	\quad
	\text{and}
	\quad
	v_2=\mathsf{T}^{-1}(w_2)
	,\]
	and we have:
	\begin{align*}
		\mathsf{T}^{-1}(\lambda w_1+w_2) & = \mathsf{T}^{-1}\left(\lambda\mathsf{T}(v_1)+\mathsf{T}(v_2)\right)\\
						 & = \mathsf{T}^{-1}\left(\mathsf{T}(\lambda v_1+v_2)\right)\\
						 & = \lambda v_1 + v_2\\
						 & = \lambda\mathsf{T}^{-1}(w_1)+\mathsf{T}^{-1}(w_2)
	\end{align*}
\end{proof}

\begin{lemma}
	Let $\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$ be an invertible transformation.
	Then:
	\[
	\mdim(\mathsf{V})<\infty\iff\mdim(\mathsf{W}) < \infty
	.\]
	In this case, $\mdim(\mathsf{V})=\mdim(\mathsf{W})$.
\end{lemma}
\begin{proof}
	Suppose that $\mathsf{V}$ is finite-dimensional.
	Let $\beta=\left\{v_1,v_2,\dotsc,v_n\right\}$ be a basis for $\mathsf{V}$.
	Then $\mathsf{T}(\beta)$ spans $R(\mathsf{T})=\mathsf{W}$, hence some subset of $\mathsf{T}(\beta)$ is a basis for $\mathsf{W}$, that is, $\mathsf{W}$ is finite-dimensional.
	The reverse implication can be proven by a similar argument using $\mathsf{T}^{-1}$.

	Now, suppose that both $\mathsf{V}$ and $\mathsf{W}$ are finite-dimensional.
	Because $\mathsf{T}$ is one-to-one and onto, we have:
	\[
	\text{nullity}(\mathsf{T}) = \mdim(N(\mathsf{T})) = \mdim(\left\{0\right\}) = 0
	,\]
	\[
	\text{rank}(\mathsf{T}) = \mdim(R(\mathsf{T})) = \mdim(\mathsf{W})
	.\]
	So by the dimension theorem,
	\[
	\text{nullity}(\mathsf{T})+\text{rank}(\mathsf{T})=\mdim(\mathsf{V})\iff 0+\mdim(\mathsf{W})=\mdim(\mathsf{V})
	,\]
	that is,
	\[
	\mdim(\mathsf{V})=\mdim(\mathsf{W})
	.\]
\end{proof}
\begin{thrm}
	Let $\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$, where $\mathsf{V}$ and $\mathsf{W}$ are finite-dimensional vector spaces with ordered bases $\beta$ and $\gamma$ respectively.
	Then:
	\[
	\mathsf{T}\ \text{invertible}\iff[\mathsf{T}]^{\gamma}_{\beta}\ \text{invertible}
	.\]
	Furthermore:
	\[
	\left[\mathsf{T}^{-1}\right]^{\beta}_{\gamma}=\left(\left[\mathsf{T}\right]^{\gamma}_{\beta}\right)^{-1}
	.\]
	
\end{thrm}
\begin{proof}
	Suppose that $\mathsf{T}$ is invertible.
	This implies that $\mdim(\mathsf{V})=\mdim(\mathsf{W})$.
	Let $n=\mdim(\mathsf{V})$.
	Then $[\mathsf{T}]^{\gamma}_{\beta}$ is an $n\times n$ matrix.
	Next, for $\mathsf{T}^{-1}:\mathsf{W}\mapsto\mathsf{V}$, we have $\mathsf{T}\circ\mathsf{T}^{-1}=I_{\mathsf{W}}$ and $\mathsf{T}^{-1}\circ\mathsf{T}=I_{\mathsf{V}}$.
	Thus:
	\[
		I_{n} = \left\{
			\begin{aligned}
				[I_{\mathsf{V}}]_{\beta} & = \left[\mathsf{T}^{-1}\circ\mathsf{T}\right]_{\beta} = \left[\mathsf{T}^{-1}\right]^{\beta}_{\gamma}[\mathsf{T}]^{\gamma}_{\beta}\\
				[I_{\mathsf{W}}]_{\gamma} & = \left[\mathsf{T}\circ\mathsf{T}^{-1}\right]_{\gamma} = [\mathsf{T}]^{\gamma}_{\beta}\left[\mathsf{T}^{-1}\right]^{\beta}_{\gamma}
			\end{aligned}
			\right.
			\implies
			[\mathsf{T}]^{\gamma}_{\beta}\ \text{invertible}
	,\]
	and
	\[
		\left([\mathsf{T}]^{\gamma}_{\beta}\right)^{-1} = \left[\mathsf{T}^{-1}\right]^{\beta}_{\gamma}
	.\]
	Now suppose that $A=[\mathsf{T}]^{\gamma}_{\beta}$ is invertible.
	Then $\exists B\in\mathsf{M}_{n\times n}$ such that $AB=BA=I_{n}$.
	We apply theorem \ref{uniq} to a starting vector space $\mathsf{W}$ with an ordered basis $\gamma=\left\{\gamma_1,\gamma_2,\dotsc,\gamma_n\right\}$ and a destination vector space $\mathsf{V}$ with the set of vectors:
	\begin{align*}
		v_1 & = b_{11}\beta_1+b_{21}\beta_2+\dotsb+b_{n1}\beta_n\\
		v_2 & = b_{12}\beta_1+b_{22}\beta_2+\dotsb+b_{n2}\beta_n\\
		    &\nvdots\\
		v_n & = b_{1n}\beta_1+b_{2n}\beta_2+\dotsb+b_{nn}\beta_n
	\end{align*}
	where $\beta=\left\{\beta_1,\beta_2,\dotsc,\beta_n\right\}$, an ordered basis for $\mathsf{V}$.
	Then $\exists \mathsf{U}\in\mathcal{L}(\mathsf{W},\mathsf{V})$, such that:
	\[
	\mathsf{U}(\gamma_k) = v_k = b_{1k}\beta_1+b_{2k}\beta_2+\dotsb+b_{nk}\beta_n,\quad k=1,2,\dotsc,n
	.\]
	If follows that $[\mathsf{U}]^{\beta}_{\gamma}=B=(b_{ik})$.
	To show $\mathsf{U}=\mathsf{T}^{-1}$, note that:
	\[
[\mathsf{U}\circ\mathsf{T}]_{\beta}=[\mathsf{U}]^{\beta}_{\gamma}[\mathsf{T}]^{\gamma}_{\beta}=BA=I_n=[I_{\mathsf{V}}]_{\beta}
	.\]
	So $\mathsf{U}\circ\mathsf{T}=I_{\mathsf{V}}$, and similarly, $\mathsf{T}\circ\mathsf{U}=I_{\mathsf{W}}$.
	Thus $\mathsf{T}$ is invertible.
\end{proof}

\begin{definition}
	Isomorphic.
	Let $\mathsf{V}$ and $\mathsf{W}$ be vector spaces.
	We say that $\mathsf{V}$ is \textbf{isomorphic} to $\mathsf{W}$ if $\exists\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$ that is \textbf{invertible}, and denote:
	\[
	\mathsf{V}\simeq\mathsf{W}
	.\]
	Such a linear transformation is called an \textbf{isomorphism} from $\mathsf{V}$ onto $\mathsf{W}$.
\end{definition}
\begin{exmp}
	The vector space $\mathsf{M}_{2\times 2}(\mathbb{R})$ is isomorphic to  $\mathbb{R}^{4}$, because the transformation
	\[
	\mathsf{T}
	\begin{bmatrix}
		a & b\\
		c & d\\
	\end{bmatrix}
	= (a,b,c,d)
	,\]
	is linear, one-to-one, and onto.
\end{exmp}
\begin{thrm}
	Let $\mathsf{V}$ and $\mathsf{W}$ be finite-dimensional vector spaces over the same field.
	Then:
	\[
	\mathsf{V}\simeq\mathsf{W}\iff\mdim(\mathsf{V})=\mdim(\mathsf{W})
	.\]
\end{thrm}
\begin{proof}
	Suppose that $\mathsf{V}\simeq\mathsf{W}$ and that $\mathsf{T}:\mathsf{V}\mapsto\mathsf{W}$ is an isomorphism from $\mathsf{V}$ onto $\mathsf{W}$.
	Then we have $\mdim(\mathsf{V})=\mdim(\mathsf{W})$.
	Now suppose that $\mdim(\mathsf{V})=\mdim(\mathsf{W})$, and let $\beta=\left\{v_1,v_2,\dotsc,v_n\right\}$ and $\gamma=\left\{w_1,w_2,\dotsc,w_n\right\}$ be bases for $\mathsf{V}$ and $\mathsf{W}$ respectively.
	By theorem \ref{uniq} we can say that $\exists\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$ such that $\mathsf{T}(v_i)=w_i$ for $i=1,2,\dotsc,n$.
	Then we have:
	\[
	R(\mathsf{T})=\mspan(\mathsf{T}(\beta)) = \mspan(\gamma)=\mathsf{W}
	.\]
	So $\mathsf{T}$ is onto.
	Then $\mathsf{T}$ is also one-to-one, and hence $\mathsf{T}$ is an isomorphism.
\end{proof}
The following theorem shows how the collection $\mathcal{L}(\mathsf{V},\mathsf{W})$ of all linear transformations my be identified with the appropriate vector space of $m\times n$ matrices.
\begin{thrm}
	Let $\mathsf{V}$ and $\mathsf{W}$ be vector spaces where:
	\[
	\mdim(\mathsf{V})=n\quad\text{and}\quad\mdim(\mathsf{W})=m
	,\]
	and where $\beta$ and $\gamma$ are the ordered bases for $\mathsf{V}$ and $\mathsf{W}$ respectively.
	Then the transformation $\Phi:\mathcal{L}(\mathsf{V},\mathsf{W})\mapsto\mathsf{M}_{m\times n}(F)$ defined by:
	\[
	\Phi(\mathsf{T})=[\mathsf{T}]^{\gamma}_{\beta},\ \forall\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})
	,\]
	is an \textbf{isomorphism}.
\end{thrm}
\begin{proof}
	By theorem \ref{linear}, $\Phi$ is linear.
	All that's left to show is that $\Phi$ is one-to-one and onto.
	This can be achieved by demonstrating that $\forall A\in\mathsf{M}_{m\times n}$, $\exists! \mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$ such that $\Phi(\mathsf{T}) = A$.
	\bigbreak
	\noindent
	Let $\beta=\left\{\beta_1,\beta_2,\dotsc,\beta_n\right\}$, $\gamma=\left\{\gamma_1,\gamma_2,\dotsc,\gamma_m\right\}$, and let $A=(a_{ij})$ be a given $m\times n$ matrix.
	Applying theorem \ref{uniq} to the starting vector space $\mathsf{V}$ with the ordered basis $\beta$ and a destination vector space $\mathsf{W}$ with the set of vectors:
	\begin{align*}
		w_1 & = a_{11}\gamma_1 + a_{21}\gamma_2 + \dotsb + a_{m1}\gamma_m\\
		w_2 & = a_{12}\gamma_1 + a_{22}\gamma_2 + \dotsb + a_{m2}\gamma_m\\
		    &\nvdots\\
		w_n & = a_{1n}\gamma_1 + a_{2n}\gamma_2 + \dotsb + a_{mn}\gamma_m
	\end{align*}
	where $\gamma=\left\{\gamma_1,\gamma_2,\dotsc,\gamma_n\right\}$ form an ordered basis for $\mathsf{W}$.
	Then there exists a unique linear transformation $\mathsf{T}:\mathsf{V}\mapsto\mathsf{W}$ such that:
	\[
	\mathsf{T}(\beta_j) = w_j = a_{1j}\gamma_1 + a_{2j}\gamma_2 + \dotsb + a_{mj}\gamma_m,\ j=1,2,\dotsc,n
	.\]
	Which implies that $[\mathsf{T}]^{\gamma}_{\beta}=A$, that is, $\Phi(\mathsf{T})=A$.
	Therefore $\Phi$ is an isomorphism.
\end{proof}

\begin{definition}
	Standard Representation.
	Let $\beta$ be an ordered basis for an $n$-dimensional vector space $\mathsf{V}$ over the field $F$.
	The \textbf{standard representation} of $\mathsf{V}$ with respect to $\beta$ is the function $\phi_{\beta}:\mathsf{V}\mapsto F^{n}$ defined by:
	\[
	\phi_\beta(v)=[v]_\beta,\ \forall v\in\mathsf{V}
	.\]
\end{definition}
\begin{exmp}
	Let $\beta=\left\{(1,0),(0,1)\right\}$ and $\gamma=\left\{(1,2),(3,4)\right\}$ be ordered bases for $\mathbb{R}^2$.
	For $v=(1,-2)$, we have
	\[
	\phi_\beta(v) = [v]_\beta=
	\begin{bmatrix}
		1\\
		-2\\
	\end{bmatrix}
	\quad
	\text{and}
	\quad
	\phi_\gamma(v) = [v]_\gamma=
	\begin{bmatrix}
		-5\\
		2\\
	\end{bmatrix}
	.\]
\end{exmp}
\begin{thrm}
	For any finite-dimensional vector space $\mathsf{V}$ with ordered basis $\beta$, $\phi_\beta$ is an isomorphism.
\end{thrm}

Now let $\mathsf{V}$ and $\mathsf{W}$ be vector spaces of dimension $n$ and $m$ respectively, and let $\mathsf{T}\in\mathcal{L}(\mathsf{V},\mathsf{W})$.
Next, define $A=[\mathsf{T}]^{\gamma}_{\beta}$, where $\beta$ and $\gamma$ are arbitrary ordered bases of $\mathsf{V}$ and $\mathsf{W}$ respectively.
Then the following \textbf{commutative diagram} maps the relationships between $\mathsf{V},\mathsf{W},F^{n}$, and $F^{m}$:

\begin{center}
	\begin{tikzpicture}[node distance=2cm, auto]
		\node (V) {$\mathsf{V}$};
		\node (W) [node distance=5cm, right of=V] {$\mathsf{W}$};
		\node (Fn) [below of=V] {$F^{n}$};
		\node (Fm) [below of=W] {$F^{m}$};
		\draw[-latex] (V) to node {$\mathsf{T}$} (W);
		\draw[-latex] (V) to node [swap] {$\phi_\beta$} (Fn);
		\draw[-latex] (W) to node {$\phi_\gamma$} (Fm);
		\draw[-latex] (Fn) to node [swap] {$\mathsf{L}_A$} (Fm);
	\end{tikzpicture}
\end{center}
It also shows that there are two composites of linear transformations that map $\mathsf{V}$ into $F^{m}$, and thus we can conclude that
\[
\mathsf{L}_A\phi_\beta = \phi_\gamma\mathsf{T}
,\]
that is, that the diagram ``commutes''.
This diagram allows us to transfer operations on abstract vector spaces to ones on $F^{n}$ and $F^{m}$.
\begin{exmp}
	Let us define $\mathsf{T}\in\mathcal{L}(P_3(\mathbb{R}),P_2(\mathbb{R}))$ by:
	\[
	\mathsf{T}(f(x)) = f^{\prime}(x)
	.\]
	Let $\beta$ and $\gamma$ be the standard ordered bases for $P_3(\mathbb{R})$ and $P_2(\mathbb{R})$ respectively, and let $\phi_\beta:P_3(\mathbb{R})\mapsto\mathbb{R}^{4}$ and $\phi_\gamma:P_2(\mathbb{R})\mapsto\mathbb{R}^{3}$ be the corresponding standard representations of $P_3(\mathbb{R})$ and $P_2(\mathbb{R})$.
	If $A=[\mathsf{T}]^{\gamma}_{\beta}$ then
	\[
	A = 
	\begin{bmatrix}
		0 & 1 & 0 & 0\\
		0 & 0 & 2 & 0\\
		0 & 0 & 0 & 3\\
	\end{bmatrix}
	.\]
	Now consider $p(x)=2+x-3x^2+5x^3$, we can see that:
	\[
	\mathsf{L}_A\phi_\beta(p(x))=
	\begin{bmatrix}
		0 & 1 & 0 & 0\\
		0 & 0 & 2 & 0\\
		0 & 0 & 0 & 3\\
	\end{bmatrix}
	\begin{bmatrix}
		2\\
		1\\
		-3\\
		5\\
	\end{bmatrix}
	=
	\begin{bmatrix}
		1\\
		-6\\
		15\\
	\end{bmatrix}
	.\]
	But since $\mathsf{T}(p(x))=p^{\prime}(x)=1-6x+15x^2$, we have:
	\[
	\phi_\gamma\mathsf{T}(p(x)) = 
	\begin{bmatrix}
		1\\
		-6\\
		15\\
	\end{bmatrix}
	.\]
\end{exmp}
