---- > [!definition] Definition. ([[vector space of m-by-n matrices]]) > The set of all $m \times n$ [[matrix|matrices]] with entries in an [[integral domain]] $R$ forms an $R$-[[module]], variously denoted $\mathcal{M}_{m,n}(R)$, $\text{Mat}_{m,n}(R)$, or — when $R=\mathbb{F}$ is a [[field]] — $\mathbb{F}^{m \times n}$. > > Basis elements multiply via the formula $E_{ij}E_{k \ell}=\delta_{jk}E_{i \ell}$. Mnemonic: "matching middles move forward". ^definition > [!specialization] Statement for $R$ a field. > For $m,n \in \mathbb{N}$, set of all $m$-by-$n$ [[matrix|matrices]] with entries in $R=\mathbb{F}$ forms a [[vector space]] over $\mathbb{F}$, denoted $\mathbb{F}^{m \times n}$. It has [[dimension]] $mn$. ^specialization > [!justification] > We will show that the [[vector space of m-by-n matrices]] is indeed a [[vector space]] over [[field]], with [[dimension]] $mn$. The work for the general module case is pretty much identical. > > #### Definition of Addition > We define the **sum** $+:(\ff^{m \times n} \times \ff^{m \times n}) \to \ff^{m \times n}$ of two identically-sized [[matrix|matrices]] to be the [[matrix]] obtained by performing component-wise addition on the entries in each [[matrix]]. > > #### Definition of Scalar Multiplication > We define the **product** between a [[matrix]] in $\ff^{m \times n}$ and a [[scalar]] in $\ff$ is the new [[matrix]] obtained by multiplying each entry by said [[scalar]]. > > ### Satisfaction of Axioms > Commutativity and additive associativity follow from the related properties of $\ff$. $\vec 0$ is the [[matrix]] of the $0$-[[linear map|map]] and is unique because the $0$-map is unique. The multiplicative identity holds too because of properties of $\ff$, so does multiplicative associativity. Additive inverses are just a matrix multiplied by the inverse of the additive identity in $\ff$. Finally, the distributive properties hold because because they do in $\ff$. > > ### On Dimension > For $1 \leq j \leq m$ and $1 \leq k \leq n$, let $\tilde \MM_{ij}$ denote the [[matrix]] with $1$ in entry $i,j$ and $0$ in each other entry. Consider the set $B=\{\tilde \MM_{i,j}\}_{i=1, j=1}^{m,n}.$ > We claim that $B$ is a [[basis]] of $\ff^{m \times n}$; note that $|B|=mn$. Clearly $B$ [[spans]] $\ff^{m \times n}$. [[linearly independent|Linear Independence]] holds as well because for $d_{11} \dots, d_{mn} \in \ff$ the [[linear combination]] $d_{11}\tilde \MM_{1,1} + \dots + d_{mn} \tilde \MM_{m,n}$ > is a [[matrix]] $M$ with the [[scalar]] $d_{ij}$ placed in index $ij$. Obviously $M=0$ if and only if each $d_{ij}$ is $0$; thus, $B$ is a [[basis]] for $\ff^{m \times n}$ with length $mn$. $\qedin$ > ---- #### ---- #### 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 > ```