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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 $A_{i,j} = 0$ ?

Gerry Myerson
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Shuzheng
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    You're exactly right. This holds in every case but characteristic 2. To see this, $A=-A\implies 2A=0$, 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. – Ian Coley Feb 05 '14 at 10:31
  • In case $\mathbb F = {[0],[1]}$ you may only need to look at the Identity matrix as a counter example. – Math137 Feb 05 '14 at 10:35
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    @IanColey Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Apr 27 '15 at 14:52

1 Answers1

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[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.

Ian Coley
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