Prove that for a vector space $V$ if $v,w\in V,\ v^TAw=O$ then $\dim V\leq m$ where $A $ is a $2m\times 2m$ matrix 
Question: Let $A$ be an invertible real $n × n$ matrix such that $A^T = −A$. Prove
that $n$ is even. Put $n = 2m$. A subspace $V\subset \mathbb{R}^n$ is said to be isotropic for $A$ if $v^TAw=O$ where $v,w\in V$. Prove that if $V$ is isotropic then $\dim V\leq m$.

Now i got the first part. Its just as $A^T=-A$ therefore $\det(A^T)=\det(-A)\implies \det(A)=(-1)^n \det(A)$. Hence $n$ has to be even.
But i am stuck about the second part. How to approach this.
 A: Let $v_1, \ldots, v_k$ be an orthogonal basis for $V$. We want to show that $k \le m$.
Now define $u_i = Av_i$.
Claim: $u_i \not\in V$.
Proof: If $u_i$ was in $V$ then $u_i^TAv_i = \|u_i\|^2 \neq 0$ (this is because $v_i \neq 0$ and $A$ invertible).
Further, due to the invertibility of $A$ and the independence of the $v_i$ we see that the $u_i$ are independent as well.
Finally, we see that
$$u_i^T v_j = v_i^TA^Tv_j = - v_i^TAv_j = 0.$$
Thus, if $U$ is the space spanned by the $u_i$. Then $U \perp V$ and $\dim U = \dim V$. However, due to $U \perp V$, we have that $\dim u + \dim V \le 2m$. Thus we get that $\dim V \le m$.
A: LEMMA
Suppose that $e_1, \dots, e_m \in \mathbb R^n$ are linearly independent, orthogonal vectors of $V$. Then $e_1, \dots, e_m, Ae_1, \dots, Ae_m$ are also linearly independent.
For the proof, suppose that
$$\sum_{i=1}^{m} \alpha_ie_i + \sum_{i=1}^{m} \beta_i Ae_i=0$$ then for $1 \le j \le m$
$$\sum_{i=1}^{m} \alpha_ie_{j}^Te_i + \sum_{i=1}^{m} \beta_i e^T_{j}Ae_i= \alpha_j \Vert e_j \Vert^2=0.$$ Which imply that $\alpha_1 = \dots = \alpha_m =0$. As $A$ is bijective ($\det A \neq 0$), $Ae_1, \dots,Ae_m$ are also linearly independent and $\beta_1 = \dots = \beta_m =0$ too, which proves the lemma.
We then get the desired result as the cardinal of a linearly independent family of vectors is less or equal to the dimension of the space.
