I'm having problem showing the following:
All operators are defined on $V$ which is real (not complex). Let $f$ be a bilinear operator that is anti-symetric (meaning $f(a,b)=-f(b,a))$ and let $J$ be a linear operator s.t $J^2=-I$. Also, I know that rank$(f)=\dim V=n$. I need to prove that if for every $a,b$ in $V$ we define $S$ bilinear operator s.t: $$ S(a,b) = f(a,J(b)) $$ then $S$ is symmetric $(S(a,b)=S(b,a))$ and positive definite $(S(v,v)\geq v$ for all $v$ in $V$).
My thoughts are: Immediately notice that $a^tSb=afJb \iff S=fJ$
However I don't know what to do from here.
(I gave some examples for myself for possible J's and it seems to be anti-symetric but I don't know how to prove it).
Thanks in advance.
None of this is true.
As darij grinberg already explained in the comments, $S$ is not necessarily positive definite. A simple example would be $f(a,b) = a^T J b$ for antisymmetric $J$ such that $J^2 = -{\rm I}$, for example, $J = \left[\begin{smallmatrix} & 1 \\ -1 \end{smallmatrix}\right]$.
As for symmetricity (order must be even if we want to have a real $J$ such that $J^2 = {\rm I}$), I'm quite sure that the statement is true for the matrices of order $2$. However, for order $4$, it is not too hard to construct a counterexample:
\begin{align*} F &:= Y - Y^T = \begin{bmatrix} 0 & -3 & -6 & -9 \\ 3 & 0 & -3 & -6 \\ 6 & 3 & 0 & -3 \\ 9 & 6 & 3 & 0 \end{bmatrix}, \quad Y := \begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \end{bmatrix}, \\ J &:= \begin{bmatrix} 1 & -2 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 2 \\ 0 & 0 & -1 & -1 \end{bmatrix}. \end{align*}
It is easy to see that $f(a,b) = a^T F b$ defines an antisymmetric bilinear form and that $J^2 = -{\rm I}$.
But,
$$S(e_1, e_4) = -3 \ne 15 = S(e_4, e_1),$$
where $e_1$ and $e_4$ denote the first and the fourth vectors of the canonical basis of $\mathbb{R}^4$.
On a side note, $J$ need not be antisymmetric. Check my example (as a whole, and/or each of the diagonal blocks in $J$).