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.
 A: I do not think that the property you want to prove depends on the characteristic. $W_1 \cap W_2 = \{0\}$ holds true for any field. As $A^t = A$ and $A_{ij} =0$ for $i \le j$ implies $A_{ji} = (A^t)_{ji} = A_{ij} = 0$ for $i \le j$, hence $A = 0$. So it suffices to show that $\dim W_1 + \dim W_2 = n^2$. $W_1$ has as a basis the $\frac{(n-1)n}2$ matrices $E_{ij}$, $i > j$, where $(E_{ij})_{kl} = \delta_{ik}\delta_{jl}$, and $W_2$ has as a basis the $n$ matrices $E_{ii}$ together with the $\frac{n(n-1)}2$ matrices $E_{ij} + E_{ji}$, $i>j$. So 
$$ \dim W_1 + \dim W_2 = (n-1)n + n = n^2. $$
That is $M_n(\mathbf F) = W_1 \oplus W_2$.

But for ${\rm char}\,\mathbf  F \ne 2$, we can write 
$$ A = \frac{A+A^t}2 + \frac{A-A^t}2 $$
And know write the antisymmetric part $\frac{A-A^t}2$ as a difference of twice its "upper part" (in $W_1$) and the symmetric matrix which has the same "upper part" as $\frac{A+A^t}2$ and the lower part the transpose of the upper part. So a more direct approach is possible if we can divide by $2$.
A: 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
