i hope my question is not too trivial.
Let's suppose we have a vector space $V$ with a unimodular quadratic form $q$ of signature $(m,n)$. 


My question is: which is the maximum dimension of an istropic subspace of $V$?


I would say it is $min\{m,n\}$. This is why. Suppose $min\{m,n\}=m$, if $min\{m,n\}=n$ the proof is the same. Given $U$, a positive subspace of dimension $m$, i choose $u\_1,\cdots, u\_m$ an ortonormal basis of $U$. Then i consider $U^\perp$ which is negative and i choose $w\_1,\cdots,w\_n$ an ortonormal basis. So the vectors $\{u\_1+w\_1,u\_2+w\_2,\cdots,u\_m+w\_m\}$ span an isotropic subspace of dimension $m$.


But could there be an isotropic subspace of dimension more than $min\{m,n\}$?



Maximum dimension of an isotropic subspace in a quadratic space
This is the Witt index of the quadratic form. I'm assuming you're working over $\bf{R}$, in which case it is indeed $\min\{m,n\}$ (use Witt cancellation). It can be smaller over other fields (like $\bf{Q}$); for instance, you could have an indefinite quaternary form which is anisotropic. For more details see Chapter 1 of Milnor and Husemoller's "Symmetric Bilinear Forms" or Lam's "Introduction to Quadratic Forms over Fields".


