Index of bilinear form

Question

I am faced with the following problem:

Given $\beta$ a bilinear form with determinant less than 0 in $R^6$ such that:

$\beta(u_1, u_1) > 0$

$\beta(u_2, u_2) < 0$

$\beta(u_4, u_4) > 0$

$\beta(u_5, u_5) < 0$

where $\{u_1, u_2, u_3, u_4, u_5, u_6\}$ is a basis of $R^6$ and that $L_1 = span(\{u_1, u_2, u_3\})$ is orthogonal to $L_2 = span(\{u_4, u_5, u_6\})$ with respect to $\beta$, prove that the index and coindex of $\beta$ are both 3.

I can't understand why this is true. Couldn't the index be 1 and the coindex 5, for example? I have, however, failed at finding a counterexample.

Answer 1

The index is the maximal possible dimension of a negative definite subspace. Since the form $\beta$ is negative definite on at least the span of $\{u_2, u_5\}$ (a 2-dimensional subspace), its index cannot be smaller than 2.

Similarly, the co-index cannot be smaller than 2. Therefore, the index cannot exceed 4.

Why can't the index equal 4? This, I think, needs to be examined using the negativity of the determinant of $\beta$.