Symmetric matrix/R/Diagonalizable/Fact/Proof

Proof

We consider the matrix ${}M$ as the Gram matrix of a symmetric bilinear form. Due to exercise, we have

${}n=p+q+a\,,$

where ${}(p,q)$ is the type of the bilinear form, and where ${}a$ is the dimension of the degeneracy space. Because of exercise, the degeneracy space is the eigen space of ${}M$ (considered as a linear mapping on ${}\mathbb {R} ^{n}$) of the eigen value ${}0$. From fact we know that ${}p$ (${}q$) is the sum of the dimensions of the eigen spaces of the positive (negative) eigen values. Therefore, ${}\mathbb {R} ^{n}$ is the sum of the eigen spaces. Hence, ${}M$ is diagonalizable.

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