Number of $1$'s in Silvester's form of bilinear form $x^tAx$ is the sum of dimensions of positive eigenvalues of $A$.

Question

I thought about the following statement, I don't know if this is true and if it is then how to prove it. Let $A$ be a symetric matrix over $\mathbb{R}$. Define the bilinear form $f(v,w) = v^tAw$. Consider the silvester form of $f$ (the one with $1$'s, $-1$'s and $0$'s on the diagnole and $0$ everywhere else).
Is it true that the number of $1$'s is the sum geometric multiplicities of positive eigenvalues of $A$ (and similarly the number of $0$'s is the geometric multiplicity of eigenvalue $0$, and the number of $-1$'s is the sum of geometric multiplicities of negative eigenvalues). For example if $A=I$ then there is only one eigenvalue, it is positive and it has a geometric multiplicity of $2$, and the silvester form really has two $1$'s on the diagnole and that's it. Will it always be the case? If so, why?