Why does the count of positive pivots from one Gaussian elimination always equal the count of positive eigenvalues, turning an expensive root-finding problem into a cheap pivot count?
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Positive pivots match positive eigenvalues in number
State the theorem for symmetric matrices, pairing the count of positive pivots with the count of positive eigenvalues, and record that pivots are not eigenvalues while both products equal the determinant.
Cost of pivots versus roots of
Write the comparison for a matrix: elimination produces fifty pivots quickly and stably, while extracting roots of the characteristic polynomial is computationally bad.
Shifting by to locate eigenvalues
Record the counting method, taking the pivots of to give the number of eigenvalues above 7, and note how repeated shifts narrow an eigenvalue down.