# Proof of Sylvester's Law of Inertia

I'm getting confused about the proof in my notes of Sylvester's Law of Inertia, and positive definiteness in general. I'll try to state my problems explicitly:

The theorem states that if $\psi : V\times V \to \mathbb{R}$ is a real symmetric bilinear form, represented by $$\left( \begin{array}{c c c} I_p \\ & -I_q & \\ & & 0 \end{array} \right) \;\;\text{and}\;\; \left( \begin{array}{c c c} I_p' \\ & -I_q' & \\ & & 0 \end{array} \right)$$

then $p = p'$ and $q = q'$.

1. The proof I have shows that $p$ is the largest possible dimension of a positive definite subspace, and so $p = p'$. Why is this implication true?

2. It then goes on to say that "similarly, $q = q'$". What exactly does this mean?

3. Wouldn't it suffice, after having shown uniqueness of $p$, to say that the rank($= p + q$) of a matrix is well-defined, and so it follows $q$ is well-defined?

4. What does it actually mean for a real symmetric bilinear form to be positive definite? The definition of positive-definitness in my notes is in terms of subspaces. Does it mean that the entire space is positive definite (w.r.t that particular form)? If so, how does this relate to matrices representing the form? How do we get from this definition to the one I'm finding online, which states that $A$ is positive definite iff $x^T A x > 0$ for all x in the space?

Thanks. I know that I'm asking a lot, and will greatly appreciate any help.

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1. If you can show that $p$ is the dimension of the largest subspace on which the form is positive definite, then the same argument shows that $p'$ is also the dimensions of the largest subspace on which the form is positive definite. Thus $p = p'$. Think of it this way, a priori $p$ seems to depend on the choice of basis, but then you show that you can also define $p$ in a basis invariant way.
3. A form $\psi$ is positive definite if $\psi(v,v) > 0$ for all $v \ne 0$. A consequence Grahm-Schmidt and Sylvester's law of interia is that a matrix $A$ represents a positive definite form if and only if $A = P Id P^t = P P^t$ for some invertible matrix $P$.