Proof of the Wirtinger inequality This is an exercise from the book "Complex Geometry, An Introduction" by Huybrechts. The statement involves to prove that the restriction of the fundamental form $\omega$ of a vector space $V, I, \langle, \rangle$ (with real dimension $2n$ and an almost complex structure $I$ compatible with the inner product $\langle, \rangle$), to a sub-space $W$ of real dimension $2m$, is not greater than $m! vol$ where $vol$ is the volume form of $W$. 
$$\omega^m|_W\le m!\cdot vol$$
The book has a hint: There exists an orthonomal basis $\{u_i, v_i:i=1,2,\cdots, m\}$ of $W$ such that $$\omega|_W=\sum_{i=1}^m \lambda_i u^i\wedge v^i$$ Where $u^i, v^i, i=1, 2, \cdots, m$ are the dual basis. From this one is easy to prove the conclusion.  
I did find a proof without using the hint from the book "Complex Analytic Sets" by E.M. Chirka and I understood the proof well. My question is how to prove the expression in the hint.  
Let $P: V\rightarrow W$ be the orthogonal projection and $J=P\circ I$ then I was led to the conclusion that the existence of the above expression depends on a fact that $J^2: W\rightarrow W$ has a real eigenvalue whose eigenspace is of dimension (at least) $2$. Then I don't know how to prove this.
Any hints are appreciated.
After reading Ted's answer I realized that $J$ is actually skew-symmetric for $\langle Jv, w\rangle=\langle Iv, w\rangle$ since $(Iv-Jv)\perp W$. Then $J$ has the desired block diagonal form.
 A: This is just linear algebra (and nothing to do with projection). It's really just the process for obtaining the normal form for a skew-symmetric bilinear form.
First of all, if $\omega|_W$ is degenerate, then the result is obvious, as $\omega^m|_W = 0$. So we suppose $\omega|_W$ is nondegenerate. This means that $\omega|_W$ is represented by a nonsingular skew-symmetric real $2m\times 2m$ matrix $A$ (here I'm assuming that $\langle\cdot,\cdot\rangle$ is a real inner product on $V$).  $\sqrt{-1}A$ is a hermitian matrix, which is unitarily diagonalizable by the Spectral Theorem, and so therefore is $A$, with pure imaginary eigenvalues. Following the usual linear algebra protocol (taking real and imaginary parts of the complex eigenvectors), we get a real normal form for $A$ with  diagonal blocks of the form $\begin{bmatrix} 0 & -\lambda_j \\ \lambda_j & 0\end{bmatrix}$ with respect to an orthonormal basis. [Note that I don't need to assume nondegeneracy. Degeneracy will manifest itself by $\lambda_j=0$ for some $j$.]
That is, we end up with an orthonormal basis $\{u_1,v_1,\dots,u_m,v_m\}$ for $W$. Letting $\{u^1,v^1,\dots,u^m,v^m\}$ denote the dual basis, this means that 
$$\omega|_W = \sum_{j=1}^m \lambda_j u^j\wedge v^j$$
for some scalars $\lambda_j$.
Remember that $\omega(u,v) = \langle Iu,v\rangle$. Then $\omega(u_j,v_j) = \lambda_j = \langle Iu_j,v_j\rangle$. By Cauchy-Schwarz,
$$|\lambda_j| = \left|\langle Iu_j,v_j\rangle\right| \le \|Iu_j\|\|v_j\| = \|u_j\|\|v_j\| = 1,$$
and you know how to finish it from here.
