# Abstract Nonsense

## Invertible Linear Transformations (Pt. II)

Point of post: This is a literal continuation of this post. Treat it as a physical continuation.

The next order of business is proving a simple but laborious set-theoretic theorem which will make our lives easier when proving that something is an isomorphism. We begin by defining some terminology. Let $X,Y$ be sets and $f:X\to Y$. We call $g:Y\to X$ a right inverse for $f$ if $f\circ g=\text{id}_Y$. Similarly, $h$ is a left inverse if $h\circ f=\text{id}_X$. With this said, we may state and prove:

Theorem: Let $X,Y$ be sets and $f:X\to Y$. Then, $f$ is bijective if and only if $f$ has a right inverse $g$ and a left inverse $h$. Moreover, if this is true then $f^{-1}=g=h$.

Proof: To prove this theorem we first prove the following claims.

Claim 1: A function possesses a left inverse if and only if it’s injective.

and

Claim 2: A function possesses a right inverse if and only if it’s surjective.

To prove this first claim we first assume that $f$ possesses a left inverse and show that this implies that $f$ is injective. To see this we merely note that if $f(x)=f(y)$ then $x=h(f(x))=h(f(y))=y$. Conversely, suppose that $f$ is injective. If $f$ is surjective then we can take $h=f^{-1}$, otherwisefix $x_0\in X$. We then define

$h:Y\to X:y\mapsto \begin{cases}x\text{ where }x\text{ is the unique element of }X\text{ such that }f(x)=y & \mbox{if}\quad y\in f(X)\\ x_0 & \mbox{if}\quad x\notin f(X)\end{cases}$

evidently then $h\circ f=\text{id}_X$.

Remark: Note that for a non-surjective function there is a left inverse for each element of $Y-f(X)$.

Now, to prove the second claim we first assume that $f$ possesses a right inverse $g$. Then, for $y\in Y$ we know that $f(g(y))=y$ so that $g(y)$ is the guaranteed element of $X$ which maps to $y$. Thus, $f$ is surjective.

Conversely, suppose that $f$ is surjective. Then, (tacitly employing the Axiom of Choice) we may select an element $x_y\in f^{-1}\left(\{y\}\right)$ for each $y\in Y$ and define

$g:Y\to X:y\mapsto x_y$

evidently then $f\circ g=\text{id}_Y$ as desired. From where the two claims follow. $\square$

Now that these claims are proven we can clearly state that $f$ is bijective if and only if it possesses left and right inverses. Now, to see that if it does possess right and left inverses $g,h$ then $g=h=f^{-1}$ we merely note that if $g\ne f^{-1}$ then there is some $y\in Y$ such that $g(y)\ne f^{-1}(y)$ and since $f$ is injective this implies that $y=f(g(y))\ne f(f^{-1}(y))=y$ which is clearly absurd. Similarly, if $h\ne f^{-1}$ then there exists some $y\in Y$ such that $h(y)\ne f^{-1}(y)$. Note though that since $f$ is surjective we have that $y=f(x)$ for some $x\in X$ and thus $x=h(f(x))=h(y)\ne f^{-1}(y)=f^{-1}(f(x))=x$ which is also absurd. The conclusion follows. $\blacksquare$
Now, as an immediate corollary of this theorem we get

Corollary: Let $T\in\text{End}\left(\mathscr{V}\right)$. Then, $T\in\text{GL}\left(\mathscr{V}\right)$ if and only if there exists $S,R\in\text{End}\left(\mathscr{V}\right)$ such that

$ST=\mathbf{1}=TR$

Remark: Note that the above corollary is true even for infinite dimensional spaces.

We now branch out in a different direction and show that, in essence, that while the above corollary is useful, for finite dimensional spaces there is a much, much simpler characterization of isomorphisms. Namely:

Theorem: Let $\mathscr{V}$ be an $n$-dimensional $F$-space. Then, for $T\in\text{End}\left(\mathscr{V}\right)$ the following are equivlent

\begin{aligned}&(1)\quad T\textit{ is a monomorphism}\\ &(2)\quad T\textit{ is an epimorphism}\\ &(3)\quad T\textit{ is an}\text{ }\textit{isomorphism}\end{aligned}

Proof: We prove this in three steps

$(1)\implies (2):$ To see this we fix a basis $\{x_1,\cdots,x_n\}$ for $\mathscr{V}$. We claim that if $T$ is a monomorphism then $\left\{T(x_1),\cdots,T(x_n)\right\}=\mathscr{B}$ is a basis for $\mathscr{V}$. To see this we first prove that $\mathscr{B}$ is linearly independent. To see this we assume that

$\alpha_1 T(x_1)+\cdots+\alpha_n T(x_n)=\bold{0}$

which by linearity implies that

$T\left(\alpha_1 x_1+\cdots+\alpha_n x_n\right)=\bold{0}$

but, by an earlier characterization of monomorphisms we know that $\ker T=\{\bold{0}\}$ and so

$\alpha_1 x_1+\cdots+\alpha_n x_n=\bold{0}$

but since $\{x_1,\cdots,x_n\}$ is a linearly independent set we may conclude that $\alpha_1=\cdots=\alpha_n=0$ and thus the linear independence of $\left\{T(x_1),\cdots,T(x_n)\right\}$ follows. Note then that this implies that

$\text{span}\left(\left\{T(x_1),\cdots,T(x_n)\right\}\right)=T\left(\text{span}\left\{x_1,\cdots,x_n\right\}\right)=T\left(\mathscr{V}\right)$

is an $n$-dimensional subspace of $\mathscr{V}$. But, by a often stated theorem this implies that $T\left(\mathscr{V}\right)=\mathscr{V}$ from where the claim follows.

Thus, trimming the fluff we see that $T$ is a monomorphism implies that $T\left(\mathscr{V}\right)=\mathscr{V}$, or said differently that $T$ is an epimorphism. $\square$

$(2)\implies (3):$ Since $T$ is already an epimorphism it suffices to show that $T$ is a monomorphism. To do this suppose that $\ker T\ne\{\bold{0}\}$. Then, there is some $x_0\in\mathscr{V}$ such that $T(x_0)=\bold{0}$. Evidently then

$T\left(\text{span}\{x_0\}\right)=\text{span}\{T(x_0)\}=\{\bold{0}\}$

Now, we may extend $\{x_0\}$ to a basis $\{x_0,x_1,\cdots,x_{n-1}\}$ for $\mathscr{V}$. Note then that

$T\left(\mathscr{V}\right)=T\left(\text{span}\{x_0,x_1,\cdots,x_{n-1}\}\right)=\text{span}\left\{T(x_0),T(x_1),\cdots,T(x_n)\right\}=\text{span}\left\{T(x_1),\cdots,T(x_n)\right\}$

where the last equality is clear from $\text{span}\left\{T(x_0)\right\}=\{\bold{0}\}$. Thus, $T\left(\mathscr{V}\right)$ is spanned by $n-1$ vectors, and thus $\dim_F T\left(\mathscr{V}\right)\leqslant n-1$. But, this implies that $T\left(\mathscr{V}\right)\subsetneq\mathscr{V}$ contrary to the assumption that $T$ is an epimorphism. It follows that $\ker T=\{\bold{0}\}$ and thus by an earlier theorem it follows that $T$ is a monomorphism. $\square$

$(3)\implies (1):$ This is trivial since an isomorphism must be a monomorphism by definition. $\square$

Thus, having proven all three implications the equivalence of the three statements follows. $\blacksquare$

Corollary: If $T\in\text{End}\left(\mathscr{V}\right)$ then $T\in\text{GL}\left(\mathscr{V}\right)$ if and only if $T$ possesses a right or left inverse which is also a linear transformation.

References:

1. Halmos, Paul R.  Finite-dimensional Vector Spaces,. New York: Springer-Verlag, 1974. Print

November 30, 2010 -

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