Symmetric Bilinear Form: index of bilinear = the number of positive eigenvalues Problem
Let $b:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$ a symmetric bilinear form and $A$ the (transformation) matrix of $b$. Further, let $\mu$ be the number of positive roots (counted by their multiplicity). Proof that $\textrm{index}(b)=\mu$.

So this intuitively makes sense after I calculated the diagonal form of a bilinear map with congruence, but I'm struggling to write the proof rigorously. What I know is:


*

*$A$ is symmetrical, therefore diagonalizable

*therefore, $\mu$ is the number of positive entries of the diagonalized matrix $\Lambda'$

*furthermore, since we are in $\mathbb{R}$ and $A$ is symmetrical, there exists a basis of $V$, so that we can diagonalize $A$ in such way, that the entries of the diagonalized matrix $\Lambda$ is either $-1$, $0$ or $1$

*according to Sylvester's theorem the number of $1$ as entries of $\Lambda$ is the same as the number of positive entries of the (normally) diagonalized matrix $\Lambda'$

*which is the definition of $\textrm{index}(b)$, so we have the desired result $\mu = \textrm{index}(b)$
I have a feeling that I can argue the third point better. Or maybe I'm totally missing the point of this exercise. Thank you for your help.

You should stress (and justify) that the eigen-values of a symmetrical matrix are always real scalars.
