Find basis such that bilinear forms attain normal form

Let

$g :=$ symmetric bilinear form : $g(v,w) := \omega(v,Jw)$ with $J \in \textrm{O}(V,g)$ and $\omega :=$ skew-symmetric bilinear form : $\omega(v,w) := g(v,Jw)$ with $J \in \textrm{Sp}(v,\omega)$.

$J$ is defined as an almost complex structure such that $J \in \textrm{O}(V,g) : J^2 = -\textrm{id}$.

The normal form of bilinear forms is defined as either $$J_{2r} = \begin{pmatrix} 0 & I_r \\ -I_r & 0 \end{pmatrix} \qquad \textrm{or} \qquad J_{2r} = \textrm{diag} \left( \begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix},...,\begin{pmatrix} 0 & 1 \\ -1 & 0\end{pmatrix} \right)$$

We also know the equality $B = A \cdot C$ for $A$ the gramian matrix associated to the basis $(v_i)$ of $g$, $B$ the matrix associated to $\omega$ and $C$ the matrix associated to $J$.

Show that there exists a basis $(v_i)$ of the $n$-dimensional $\mathbb R$ vector space $V$ such that the gramian matrices for the non degenerate symmetric bilinear form $g$ and the symplectic form $\omega(,) := g( ,J())$ have normal form and show that $g(J(), )$ defines a symplectic structure as well. What is the relationship of $g(J(), )$ to $\omega$ ?

So $g$ and $\omega$ are inverse to each other and since $g$ is non degenerate, $\omega$ is also non degenerate (follows from small lemma).

To show that $g(J(), )$ also defines a symplectic structure, I tried to show that $g(Jv,w) = -g(Jw,v)$:

$$g(Jv,w) \overset{\text{J orthogonal}}{=} g(J^2v, Jw) = g(-v, Jw) = -g(v, Jw) \overset{\text{g symmetric}}{=} -g(Jw, v).$$

So $g$ is alternating and since it is non degenerate by assumption it is symplectic, but how about showing the existence of the basis? Can anybody help???

• These trivial edits to continually have this question on the front page are not appropriate. Stop them please. – davidlowryduda May 6 '14 at 3:15
• My post does not get attention otherwise. – sj134 May 6 '14 at 12:16

STEP 1: Show that there exists a pair of vectors, $v_1$ and $v_2$, such that $g(v_1,v_1) = \pm1$, $g(v_2,v_2) = \pm1$, $g(v_1,v_2) = g(v_2,v_1) = 0$, and $Jv_2 = v_1$ (so that $\omega(v_1,v_2) = g(v_1,v_1) = 1$).
STEP 2: Let $W \subset V$ be the orthogonal complement of $span(v_1,v_2)$. Show that $J$ preserves $W$, and $g\mid_W$ is nondegenerate (which also implies $\omega \mid_W$ is non degenerate).
Then the result follows: By induction there is a basis for $W$ with respect to which $g$ and $\omega$ have normal form, and by adding $v_1,v_2$ to this basis you get a basis for $V$ for which $g$ and $\omega$ have normal form.
PROOF OF STEP 1: Pick any vector $v_1$ such that $g(v_1,v_1) \neq 0$ (the existence of such a vector follows by the nondegeneracy of $g$), by scaling we can assume $g(v_1,v_1) = \pm 1$. Let $v_2 = -Jv_1$, then $Jv_2 = -J^{2}v_1 = v_1$. Since $J$ is orthogonal, $g(v_2,v_2) = g(-Jv_1,-Jv_1) = \pm 1$. Also, $g(v_1,v_2) = g(v_1,Jv_1) = \omega(v_1,v_1) = 0$, since $\omega$ is antisymmetric.
PROOF OF STEP 2: If $g(w,v_1) = g(w,v_2) = 0$, then $g(Jw,v_1) = g(Jw,Jv_2) = g(w,v_2) = 0$, and similarly $g(Jw,v_2) = g(Jw,-Jv_1) = -g(w,v_1) = 0$. So $J$ preserves $W$. To show $g\mid_W$ is nondegenerate, pick $w \in W$. There exists $w'$ such that $g(w,w') \neq 0$. Let $\bar{w}'$ be the orthogonal projection of $w'$ onto $W$, so that $w' - \bar{w}' \in span(v_1,v_2)$. Then $g(w,\bar{w}') = g(w,\bar{w}' - w' + w') = g(w,w') - g(w,w' - \bar{w}') = g(w,w') \neq 0$.