I'm not sure if this notation is unique to my book but I'm just going to write it down anyways:
- The set of all $m \times n$ matrices with entries from a field $F$ is a vector space, and is denoted as $M_{m \times n}(F)$.
- Here, $\oplus$ means that one matrix is the direct sum of the other two.
Also just note that $W_1$ below is just the set of all $m \times n$ upper triangular matrices.
Proposition: In $M_{m \times n}(F)$ define $W_1 = \{ A \in M_{m \times n}(F) \mid A_{ij}=0\space\text{ if }\space i>j \}$ and $W_2 = \{ A \in M_{m \times n}(F) \mid A_{ij}=0 \space\text{ if }\space i \leq j\}$. Then $M_{m \times n}(F) = W_1 \oplus W_2$.
Proof: In order to show that the two sets are equal we must show that they are subsets of each other. So we begin by showing that $W_1 + W_2 \subseteq M_{m \times n}(F)$.
Suppose that $A \in W_1 + W_2$, from here we aim to show that $A \in M_{m \times n}(F)$. Then we have $A=B+C$ where $B \in W_1$ and $C \in W_2$. We also have $B_{ij}=0$ when $i > j$ and $C_{ij} = 0$ when $i \leq j$.
The sum of $B_{ij}$ and $C_{ij}$ will depend on whether $i > j$. In either case, it will be $B_{ij}$ or $C_{ij}$ (the other will become $0$), which are both members of $F$. Thus $A_{ij}=B_{ij}+C_{ij}$ will be a member of $F$. Therefore $A$ is an $m \times n$ matrix with entries from $F$, so $A \in M_{m \times n}(F)$. Thus $W_1 + W_2 \subseteq M_{m \times n}(F)$.
Now we show that $M_{m \times n}(F) \subseteq W_1 + W_2$. We will represent some matrix $A \in M_{m \times n}(F)$ as the sum of two matrices; one upper triangular matrix and one lower triangular matrix. Let $B$ be the matrix where $A_{ij}=B_{ij}$ if $i \leq j$, and $0$ elsewhere. Similarly, let $C$ be the matrix where $A_{ij}=C_{ij}$ if $i > j$, and $0$ elsewhere. Notice that $B \in W_1$ and $C \in W_2$. If $i > j$ then $$A_{ij} = C_{ij} = 0 + C_{ij} = B_{ij}+C{ij}.$$ Similarly, if $i \leq j$ then $$A_{ij} = B_{ij}=B_{ij}+0=B_{ij}+C_{ij}.$$ Both of these cases cover all $i,j$ and we see that $A_{ij}=B_{ij}+C_{ij}$ in both. Therefore $A=B+C$. But $B \in W_1$ and $C \in W_2$ so $A \in W_1 + W_2$. Notice that since there exists no matrix $D$ other than $0$ such that $D \in W_1$ and $D \in W_2$, then $W_1 \cap W_2 = \{ 0 \}$. Therefore $M_{m \times n}(F) \subseteq W_1 + W_2$.
Since we have shown that the sets are subsets of each other, then they must be equal. So by definition, $M_{m \times n}(F) = W_1 \oplus W_2$. QED.
I want some feedback, do some parts feel vague or unclear? Do some parts need further justification? Is this proof too long? Is it even right? etc.