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 $0$.
Best Answer
Paraphrasing from Strang, "Linear Algebra and Applications", 3ed, Theorem 6G, real symmetric matrix have same number of positive, negative and zero eigenvalues and pivots. Matrix need not be invertible.
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.