Prove if $A \in Mat_{n,n}(\mathbb F)$ is both symmetric and skew-symmetric then $A=0$ Prove if $A \in Mat_{n,n}(\mathbb F)$ is both symmetric and skew-symmetric then $A=0$
I know $A^T = A = -A \Rightarrow A = -A \Rightarrow A_{i,j} = -A_{i,j}$. 
Since $\mathbb F$ is a field we have $2A_{i,j} = 0 \Rightarrow 2 = 0 \lor A_{i,j} = 0$.
However how can I verify $A_{i,j} = 0$ ? Suppose $\mathbb F = \{[0],[1]\}$. Then $2 = 0$, so I cannot conclude $A_{i,j} = 0$ ?

[from my comment]
You're exactly right. This holds in every case but characteristic $2$. To see this, $A=−A\implies2A=0$, and that is true for every matrix in characteristic $2$. All you need is an explicit example to prove that a nonzero $A$ exists, e.g. the identity.
