Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Knowing that the number of the signs of pivots of a matrix is equal to the signs of eigenvalues of a matrix, that is the number of positive/negative pivots is the same the as the number of positive/negative eigenvalues, how does this help or simplify the computation to find the eigenvalues of a large matrix? It is said that it would and is a helpful fact to know but I don't understand how is this going to help.

I still have to go through the determinant and polynomial method to find the roots, wouldn't I?

share|cite|improve this question
up vote 1 down vote accepted

I'm assuming you're talking about the use of the $\mathbf L\mathbf D\mathbf L^T$ decomposition of a symmetric matrix, since it only makes sense to speak of "signs of eigenvalues" when all the eigenvalues are real like in the symmetric case.

The reason why this works out is due to the inertia theorem of Sylvester: briefly put, for a symmetric matrix $\mathbf A$ and some nonsingular matrix $\mathbf W$, $\mathbf A$ and $\mathbf W\mathbf A\mathbf W^T$ have the same number of positive, negative, and zero eigenvalues. ($\mathbf A$ and $\mathbf W\mathbf A\mathbf W^T$ are then said to be congruent.)

Since the relation $\mathbf A=\mathbf L\mathbf D\mathbf L^T$ is a congruence relation, one can then inspect the signs of the diagonal entries (pivots) of $\mathbf D$ to determine the inertia of $\mathbf A$. If the count of signs you obtained for $\mathbf D$ and for the eigenvalues of $\mathbf A$ don't match, then you've made an error somewhere.

I still have to go through the determinant and polynomial method to find the roots, wouldn't I?

For hand computations, yes. One efficient method used by computers involves performing a similarity transformation of the original symmetric matrix into a tridiagonal matrix, a matrix whose nonzero entries are on the diagonal, subdiagonal, and superdiagonal. There is then a simple recursion that can be used for computing the characteristic polynomial (and this recursion can in fact be shown to be equivalent to performing an $\mathbf L\mathbf D\mathbf L^T$ decomposition of the tridiagonal matrix), and any number of methods for finding the roots of a polynomial are then available.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.