# The number of positive eigenvalues of a non-singular symmetric matrix is equal to the number of positive pivots(Strang)

I am reading "Introduction to Linear Algebra 4th Edition" by Gilbert Strang.

In this book, there is a theorem which says the number of positive eigenvalues of a non-singular symmetric matrix is the number of positive pivots.

Strang's proof of this theorem uses the intermediate value theorem.
Is there a purely algebraic proof of this theorem?

Strang assumes non-singularity.
Are the following statements true or not? :

The number of positive eigenvalues of a symmetric matrix is equal to the number of positive pivots.
The number of negative eigenvalues of a symmetric matrix is equal to the number of negative pivots.
The number of eigenvalues which are $$0$$ of a symmetric matrix is equal to the number of pivots which are $$0$$.

• Sylvester's law of inertia might be of some help here.
– Mick
Commented Apr 12, 2020 at 8:56

Let matrix A be real symmetric matrix and $$B=PAP^T$$, where P is permutation matrix. P is applied to swap rows in A such that pivot are in correct row and Gaussian elimination can proceed. Then $$PAP^T=B=LDU=U^TDU$$. L is invertible lower tri-angular matrix. U is invertible upper tri-angular matrix. D is diagonal matrix. If A is singular then at least one diagonal entry in D is 0. $$L=U^T$$ because B is symmetric. By Law of Inertia, D preserve B eigenvalue sign and vice versa. Also because D is diagonal, its pivot equal its eigenvalue. Thus for B, sign of pivot equal sign of eigenvalue. Because all permutation matrix is invertible, by applying Law of Inertia again, B preserve A eigenvalue sign and vice versa. B and A also have same pivot (A may have pivot in wrong row, thus potentially requiring permutation matrix). Thus A have same number of positive, negative and zero eigenvalue and pivot.