[proofplan] We prove that the standard matrix units satisfy the two defining properties of a basis. First, every matrix is expanded uniquely as the sum of its entries times the corresponding matrix units, which proves spanning. Second, comparing individual entries in a vanishing linear combination forces every coefficient to be zero, which proves [linear independence](/page/Linear%20Independence). The dimension statement then follows by counting the $mn$ ordered pairs of row and column indices. [/proofplan] [step:Expand an arbitrary matrix in standard matrix units] Let $A\in M_{m\times n}(k)$ be arbitrary, and write $A=(a_{rs})$, where $a_{rs}\in k$ denotes the $(r,s)$-entry of $A$ for $1\le r\le m$ and $1\le s\le n$. Define \begin{align*} B:=\sum_{i=1}^{m}\sum_{j=1}^{n} a_{ij}E_{ij}\in M_{m\times n}(k). \end{align*} We compare entries. For fixed indices $r,s$ with $1\le r\le m$ and $1\le s\le n$, the $(r,s)$-entry of $B$ is \begin{align*} B_{rs}=\sum_{i=1}^{m}\sum_{j=1}^{n} a_{ij}(E_{ij})_{rs}=a_{rs}, \end{align*} because $(E_{ij})_{rs}=1_k$ exactly when $(i,j)=(r,s)$ and is $0_k$ otherwise. Thus $B$ and $A$ have the same entries in every position, so $B=A$. Therefore the family $\{E_{ij}:1\le i\le m,\ 1\le j\le n\}$ spans $M_{m\times n}(k)$ over $k$. [guided] To prove spanning, we must show that an arbitrary matrix can be written as a finite $k$-linear combination of the matrices $E_{ij}$. Let $A\in M_{m\times n}(k)$ be arbitrary. For each row index $r$ and column index $s$, write $a_{rs}\in k$ for the $(r,s)$-entry of $A$, so $A=(a_{rs})$. The natural candidate expansion is obtained by placing each entry $a_{ij}$ in front of the matrix unit that occupies the same position. Define \begin{align*} B:=\sum_{i=1}^{m}\sum_{j=1}^{n} a_{ij}E_{ij}\in M_{m\times n}(k). \end{align*} This is a finite sum in the [vector space](/page/Vector%20Space) $M_{m\times n}(k)$, so it is a well-defined matrix. We now verify that $B=A$ by comparing entries. Fix indices $r,s$ with $1\le r\le m$ and $1\le s\le n$. The $(r,s)$-entry of $B$ is computed using entrywise addition and scalar multiplication of matrices: \begin{align*} B_{rs}=\sum_{i=1}^{m}\sum_{j=1}^{n} a_{ij}(E_{ij})_{rs}. \end{align*} By definition of the standard matrix unit, $(E_{ij})_{rs}=1_k$ when $(i,j)=(r,s)$ and $(E_{ij})_{rs}=0_k$ for every other pair $(i,j)$. Hence every term in the double sum is zero except the term with $i=r$ and $j=s$, and that term is $a_{rs}1_k=a_{rs}$. Therefore \begin{align*} B_{rs}=a_{rs}. \end{align*} Since this equality holds for every entry position $(r,s)$, the matrices $B$ and $A$ are equal. Thus every matrix $A\in M_{m\times n}(k)$ lies in the span of the family $\{E_{ij}:1\le i\le m,\ 1\le j\le n\}$. [/guided] [/step] [step:Compare entries in a zero linear combination] Suppose that coefficients $c_{ij}\in k$ satisfy \begin{align*} \sum_{i=1}^{m}\sum_{j=1}^{n} c_{ij}E_{ij}=0, \end{align*} where $0\in M_{m\times n}(k)$ denotes the zero matrix. Fix indices $r,s$ with $1\le r\le m$ and $1\le s\le n$. Taking the $(r,s)$-entry of both sides gives \begin{align*} 0=\sum_{i=1}^{m}\sum_{j=1}^{n} c_{ij}(E_{ij})_{rs}=c_{rs}. \end{align*} Thus $c_{rs}=0$ for every pair $(r,s)$. Hence the family $\{E_{ij}:1\le i\le m,\ 1\le j\le n\}$ is linearly independent over $k$. [/step] [step:Count the basis elements to obtain the dimension] The previous two steps show that $\{E_{ij}:1\le i\le m,\ 1\le j\le n\}$ is both spanning and linearly independent, hence it is a basis of $M_{m\times n}(k)$ over $k$. The index set \begin{align*} \{(i,j)\in\mathbb{N}\times\mathbb{N}:1\le i\le m,\ 1\le j\le n\} \end{align*} has $mn$ elements, since there are $m$ choices for $i$ and $n$ choices for $j$. Therefore the basis has $mn$ elements, and by the definition of dimension of a finite-dimensional vector space, \begin{align*} \dim_k M_{m\times n}(k)=mn. \end{align*} This proves the theorem. [/step]