Dimension of totally isotropic subspaces for a given quadratic form

Let $V$ be a maximal totally isotropic subspace. Let $W$ be a subspace of dimension $\max\{r,s\}$ on which $Q$ is positive or negative definite. (For instance, if $\max\{r,s\}=r$, let $W$ be the span of $e_1,\ldots,e_r$, and if $\max\{r,s\}=s$, let $W$ be the span of $e_{r+1},\ldots, e_s$.) Then $$\dim(V\cap W) = \dim(V) + \dim(W) - \dim(V+W)$$ is a basic result from linear algebra. Since $V$ is totally isotropic, $V\cap W=0$, so we get \begin{align*} \dim(V) &= \dim(V+W)-\dim(W) \\ &\leq \dim(X)-\max\{r,s\} \\ &= r + s - \max\{r,s\} \\ &= \min\{r,s\}. \end{align*} On the other hand, certainly we have $\dim(V)\geq\min\{r,s\}$, because $V$ is a maximal totally isotropic subspace and we know that there exists a totally isotropic subspace with dimension $\min\{r,s\}$.

Therefore $\dim(V)=\min\{r,s\}$.

Recall that symmetric bilinear forms over $\Bbb R$ or over any real closed field are parametrized by their signature and their rank. In particular non-degenerate forms are parametrized by their signature. The common dimension of every maximal isotropic subspace is called the Witt index of your quadratic form $Q$, and usually denoted by $\nu(Q)$. In fact

Claim For a non-degenerate quadratic form $Q$ over $\Bbb R$, $\nu(Q)=\frac 1 2(n-|s(Q)|)$ where $s(Q)$ is the signature of the associated bilinear form of $Q$.