Proving direct sum when field is NOT of characteristic $2$. Let $\mathbb{F}$ be a field that is not of characteristic $2$. Define $W_1 = \{ A \in M_{n \times n} (\mathbb{F}) : A_{ij} = 0$ whenever $i \leq j\}$ and $W_2$ to be the set of all symmetric $n \times n$ matrices with entries from $\mathbb{F}$. Both $W_1$ and $W_2$ are subspaces of $M_{n \times n} (\mathbb{F})$. Prove that $M_{n \times n} (\mathbb{F})=W_1 \oplus W_2$.
I know how to prove that $M_{n \times n} (\mathbb{F})=W_1 \oplus W_2$ for any arbitrary field $\mathbb{F}$ (by showing $W_1 \cap W_2 = \{0\}$ and any element of $M_{n \times n} (\mathbb{F})$ is sum of elements of $W_1$ and $W_2$).
But what does "$\mathbb{F}$ be a field that is NOT of characteristic $2$" mean here (I am aware of definition of characteristic of a field), i.e. how does the characteristic of field affect the elements in $M_{n \times n} (\mathbb{F})$? And how does it change the proof?
Thanks.

When you take the field $F$ to be of characteristic $2$ i.e if the field is $\mathbb Z_2$ i.e the matrix is over $\mathbb Z_2$ do you see that the problem has nothing left in it?
Hint: Try to write down a matrix over $\mathbb Z_2$ and see what happens
