Abstract Nonsense

Crushing one theorem at a time

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.


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


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


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


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


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.


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


November 30, 2010 - Posted by | Algebra, Halmos, Linear Algebra, Uncategorized | , , , , , ,


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