Let $\mathbb{K}$ be a field with characteristic zero, $n \in \mathbb{N}$ and $A \in \cal{M}_{n}\mathbb{(K)}.$ I want to prove that $A$ can be written as a linear combination of $n \times n$ matrices with all the entries zero except one of them which is one.
For example, $\begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} = 1 \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + 2 \begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} + 3 \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} + 4 \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}.$
For that I defined the matrix $M_{ab} = (m_{ij})_{n}$ where ${m_{ab}}_{ij} = \begin{cases} 1, & \text{if } i = a, j = b \\ 0, & \text{otherwise} \end{cases}$ (with $a,b \in \mathbb{N}$ such that $1 \leq a,b \leq n$).
Then we that
$$\begin{align} A & = (a_{ij})_{n} \\ & = (a_{ij} \cdot 1)_{n} & \text{($1$ is the identity for multiplication in $\mathbb{K}$)}\\ & = (a_{ij} \cdot {m_{ij}}_{ij})_{n} & \text{(by definition of $M_{ab}$)}\\ & = \sum_{1 \leq i,j \leq n} {a_{ij}M_{ij}} \end{align}$$
I just don’t know how I can justify the last step. Do you have any ideas? Or any other approaches to prove this?
Thank you so much!
You may consider the $ij^\text{th}$ entry of $B=A-\sum_{x,y}a_{xy}M_{xy}$.$$b_{ij}=a_{ij}-\sum_{x,y}a_{xy}[M_{xy}]_{ij}=a_{ij}-a_{ij}\times1=0$$since $[M_{xy}]_{ij}=\begin{cases}1,&x=i,y=j\\0,&\text{otherwise}\end{cases}$.