Q: Is every skew-symmetric matrix congruent to a diagonal matrix? Question
Prove/disprove: if A, a matrix nxn over field F is skew-symmetric then A congruents with a diagonal matrix. 
My thoughts
I know that any symmetric matrix whose entries are real can be diagonalized by an orthogonal matrix. But is it true for skew-symmetric as well? 
I really have no idea what to do here... 
I know A consists of zeros on it's diagonal. it means it's nilpotent? if so, the eigenvalues are only zeros, I guess it means that its diagonal matrix should be all zeros? 
I'm really confused here,
any lead would help, 
many thanks.

A: Hint. Suppose $\operatorname{char}(\mathbb F)\ne2$. If $A$ is skew-symmetric and $A=P^TDP$ for some invertible matrix $P$ and diagonal matrix $D$, then $D$ has to be skew-symmetric too.
When $\operatorname{char}(\mathbb F)=2$, consider $A=\pmatrix{0&1\\ 1&0}$ and also $A=I_2$.

A: Let T: $R^2\rightarrow R^2$ be a linear operator such that $T(x,y)=(-y, x)...$ let A be the matrix of T... Then A is a real skew-symmetric matrix and A is not diagonalizable over R.(Though A is diagonalizable over C)... So, skew-symmetric matrices are not diagonalizable always...
