----- > [!proposition] Proposition. ([[hom and mat are isomorphic]]) > Let $R$ be an [[integral domain]]. Recall that [[homsets in R-mod are R-modules]], as is the set of [[vector space of m-by-n matrices|m-by-n matrices]] with entries in $R$. In fact, given two [[submodule generated by a subset|finitely generated]] [[free module|free modules]] — [[the rank theorem for free modules|WLOG]] $R^{\oplus n}$ and $R^{\oplus m}$ — we have an [[module isomorphism|isomorphism]] of [[module|modules]] $\mathcal{M}_{m,n}(R) \cong \text{Hom}_{R\text{-}\mathsf{Mod}}(R^{\oplus n}, R^{\oplus m})$ given by the map $\mathcal{M}$ which sends a [[linear map]] to its [[matrix]] wrt (say) the standard basis. ^proposition > [!specialization] Statement for finite-dimensional vector spaces. > Let $V,W$ be [[vector space#Finite-Dimensional Vector Space|finite-dimensional]] [[vector space]]s, [[every vector space has a basis|with]] $\lb v_i\rb_{i=1}^n$ a [[basis]] of $V$ and $\lb w_i\rb_{i=1}^n$ a [[basis]] of $W$. > Then the map $\MM: \text{Hom}_{k\text{-}\mathsf{Vect}}(V,W) \to \ff^{m \times n}$ which takes $T$ in the [[vector space of linear maps between two vector spaces]] to its corresponding [[matrix]] with respect to $\lb v_i\rb_{i=1}^n$, $\lb w_i\rb_{i=1}^n$ is a [[linear isomorphism]] between the [[vector space of linear maps between two vector spaces]] and the [[vector space of m-by-n matrices]]. ^specialization > [!proof] > Will prove for vector spaces (the module case should be essentially the same). > > We have [[matrix of sum of linear maps is sum of matrices]] and [[matrix of scaled linear map is scaled matrix]] for arbitrary $S,T \in$ [[vector space of linear maps between two vector spaces]] and $\lambda \in$ [[field]], hence $\MM$ is [[linear map|linear]]. > > [[linear invertibility is equivalent to injectivity and surjectivity|It suffices to show that]] $\MM$ is [[surjection|surjective]] and [[injection|injective]]. Let $T \in$ [[kernel of a linear map]] $\MM$, so that $\MM(T)$ is the $0$-[[matrix]]. This implies that $[Tv_1]_{\lb w_i\rb_{i=1}^n} = \dots = [Tv_n]_{\lb w_i\rb_{i=1}^n}=(0,\dots,0)$, meaning that $Tv_k=0$ for $1 \leq k \leq n$. Because $v_1, \dots, v_n$ form a [[basis]] of $V$, for any $v \in V$ we then have $T(v)=0$ since $T(v)=T(a_1v_1 + \dots + a_nv_n) = a_1T(v_1)+ \dots + a_nT(v_n)=0.$ > Hence $T$ is the $0-$map. So $\MM$ [[linear map is injective iff kernel is trivial|is injective]]. > > Suppose $A \in$ [[vector space of m-by-n matrices]]. Let $T$ be the [[linear map]] from $V$ to $W$ representing it, i.e., such that $Tv_k = \sum_{j=1}^m A_{j,k}w_j$ > for $1 \leq k\leq n$. Obviously $\MM(T)=A$, and so $\MM$ is a [[surjection]]. We're done by [[linear invertibility is equivalent to injectivity and surjectivity]]. $\qedin$ > [!proof]- Proof. ([[hom and mat are isomorphic]]) > ~ ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```