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). “a map is invertable if there is an inverse”: 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 “nonsingular matrix”. If it doesn’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:

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 “revert” 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.