TITLE: Why is the set of Hermitian matrices with repeated eigenvalue of measure zero?
QUESTION [14 upvotes]: The Hermitian matrices form a real vector space where we have a Lebesgue measure. In the set of Hermitian matrices with Lebesgue measure, how does it follow that the set of Hermitian matrices with repeated eigenvalue is of measure zero?
This result feels extremely natural but I do not see an immediate argument for it.

REPLY [6 votes]: Here is another argument.
The idea is that in any neighbourhood of a matrix with repeated roots, there is one with distinct roots.  i.e. we are arguing that matrices with repeated roots are residual, that the matrices with distinct roots are open and dense in the space of matrices.
The idea is to put the matrix into its Jordan form.   Then you can take a small perturbation that makes all the diagonal entries distinct (only perturbing the diagonal entries).   In this form, you can compute the characteristic polynomial and you see the roots are all distinct (as they are the diagonal entries).
I suppose this is essentially Mizar's argument but in a coordinate system where there is a little less work to do.