---
title: "invertability"
type: concept
related: [Surjectivity, Product Of Linear Maps, Inverses, Basis Of Domain, Linear Map]
source: https://www.jemoka.com/posts/kbhinvertability/
confidence: high
status: active
---

A Linear Map is invertable if it can be undone. It is called a nonsingular matrix
constituents A linear map \(T \in \mathcal{L}(V,W)\)
requirements A Linear Map \(T \in \mathcal{L}(V,W)\) is called invertable if \(\exists T^{-1} \in \mathcal{L}(W,V): T^{-1}T=I \in \mathcal{L}(V), TT^{-1} = I \in \mathcal{L}(W)\).
&ldquo;a map is invertable if there is an inverse&rdquo;: that combining the commutable inverse and itself will result in the identity map.
additional information matrix invertability Matrices whose determinants are not \(0\) (i.e. it is invertable) is called &ldquo;nonsingular matrix&rdquo;. If it doesn&rsquo;t have an inverse, it is called a singular matrix.
linear map inverse is unique An invertable Linear Map has an unique inverse:
Proof:
Suppose \(T \in \mathcal{L}(V,W)\), and \(\exists S_1, S_2\) which are both inverses of \(T\). We desire \(S_1=S_2\).
So:
\begin{equation} S_1 = S_1(TS_2) = (S_1T)S_2 = IS_{2} = S_2 \end{equation}
given Product of Linear Maps is associative.
\(S_1=S_2\), as desired. \(\blacksquare\)
injectivity and surjectivity implies invertability Suppose \(T \in \mathcal{L}(V,W)\); we desire that \(T\) is invertable IFF it is both injective and surjective.
First, suppose \(T\) is invertible; that is, \(\exists T^{-1}: T^{-1}T=I, TT^{-1}=I\) We desire that \(T\) is both injective and surjective.
Injectivity: Suppose \(Tv=Tu\); we desire \(u=v\). \(u = T^{-1}(Tu) = T^{-1}(Tv) = v\) . We essentially to use the fact that \(T^{-1}\) is a function to &ldquo;revert&rdquo; the map of \(T\); as \(T^{-1}\) is a map, we know it has to revert to the same result.
Surjectivity: Recall \(T: V\to W\). WLOG let \(w \in W\), \(w=T(T^{-1}w)\). Therefore, all \(w\) is in range of \(T\).
Second, suppose \(T\) is both injective and surjective. Define a transition \(S\) such that \(T(Sw) = w\) for all \(w \in W\) (i.e. it hits just the right element to hit \(w\) as an input of \(T\).) This is made possible because \(T\) is surjective (because you can hit all \(W\)) and injective (which makes \(S\) not need to hit two different things or have two non-equal things accidentally map to the same thing.)
Evidently, \(T(Sw)=w \forall w \in W \implies (TS) = I\) by definition.
We now desire \(ST = I\). We have \((TSTv) = (TS)(Tv) = ITv = Tv\) by associativity of map multiplication. Now, \((TSTv) = Tv \implies T(ST)v = Tv\) by associativity again. This implies that \((ST)v=v\) again because \(T\) is injective: so the same input will not produce two unique outputs.
We then can show \(S\) is a linear map in the usual way.
Having constructed the desired result, \(\blacksquare\)
Alternate Proof for Finite Dimensional \(T\) So given map to bigger space is not surjective and map to smaller space is not injective, we have that the dimension of \(W = V\), we leverage the basis of each and build the using the basis of domain.
