# Invariants of eigenvalues and Sylvester's law of inertia

I have some trouble understanding Sylvester's law of inertia:

It is known that eigenvalues are invariant under a change of basis, due to the invariance of the characteristic polynomial. Sylvester's law of inertia states that for symetric bilinear forms, the amount of positive and negative elements of the diagonal matrix representing the bilinear form $s$ is invariant under a change of basis ($B=S^TAS$).

Why is only the count of positive and negative elements invariant and not the numbers themselves (I want to say eigenvalues, but eigenvalues do not really maake sense in this context)

So like in the case of Gram–Schmidt you can always rescale vectors to change the norm to one, in this case you can rescale vectors to change the value of $n=A(v,v)\neq 0$, just define $v'=v/|n|^{1/2}$ then $A(v',v')=\operatorname{sgn}(n)$. This value is $\pm 1$, and then you can do something similar to change it to whichever positive or negative value you wish preserving the sign.
$$A=\begin{pmatrix}1& 0 & \ldots & \ldots & \ldots & \ldots & \ldots\\ 0 & 1 & 0 &\ldots & \ldots & \ldots & \ldots\\ \vdots &\vdots &\ddots &\ddots & \ldots & \ldots & \ldots\\ \vdots &\vdots &0 &-1 & 0 & \ldots & \ldots \\ \vdots &\vdots &\ddots &0 & -1 &0 & \ldots\\ \vdots &\vdots &\ddots &\ddots & \ldots & \ldots & \ldots\\ \vdots &\vdots &\ddots & \ldots &0 & 0 &0 \end{pmatrix}$$
So there's some vectors whose signature is $+1$, some whose is $-1$ and some whose is $0$, unless the symmetric bilinear form is also assumed to be non-degenerate in which case there aren't any vectors such that $A(v,v)=0$ is true.