MATH 3210 Class Notes 02-21-2025

notevector-spaces

Invertibility and Isomorphisms

Defined Inverse (Linear Map)

Theorem

An invertible linear map has a unique inverse.

Notation

If $T\in \mathcal{L}(V,W)$ is invertible, let $T^{-1}\in \mathcal{L}(W,V)$ denote the inverse of $T$.

Theorem

A linear map $T$ is invertible if and only if it is injective and surjective.

Proof

("$⟹$") Assume $T\in \mathcal{L}(V,W)$ is invertible. Suppose $u,v\in V$ and $Tu=Tv$.

$$\begin{array}{rl}Tu=Tv& ⟹T^{-1}(Tu)=T^{-1}(Tv)\\ & ⟹Iu=Iv\\ & ⟹u=v⟹\text{ injective}.\end{array}$$

Let $w\in W$. Then,

$$\begin{array}{rl}w=Iw& =(T{T}^{-1})w\\ & =T(T^{-1}w)\in V⟹\text{ surjective}.\end{array}$$

("$⟸$") Assume $T$ is injective and surjective. For each $w\in W$, let $S(w)\in V$ be the unique element of $V$, such that $T(S(w))=w$. This is guaranteed to exist and be unique since $T$ is injective and surjective. Thus, $TS=I_w$. Also, let $v\in V$, then $T(ST(v))=(TS)(Tv)=I_w(Tv)=Tv$. Therefore, since $T$ is injective, $v=STv$, so $ST=I_v$.

Finally, we must show that $S$ is a linear map. Let $w_1,w_2\in W$ and $\lambda \in \mathbb{F}$, then $T(S(w_1)+S(w_2))=TS{w}_1+TS{w}_2=w_1+w_2$. Thus, $S{w}_1+S{w}_2$ is the unique element of $V$ which maps to $w_1+w_2$ under $T$. Therefore, $S(w_1+w_2)=S(w_1)+S(w_2)$ by definition.

Question

Is injectivity alone (or surjectivity alone) enough to have invertibility of a linear map $T\in \mathcal{L}(V)$?

Answer

No, if $V$ is infinite-dimensional; yes, if $V$ is finite-dimensional.