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.



Why is the set of Hermitian matrices with repeated eigenvalue of measure zero?
Call $S$ the set of matrices with repeated eigenvalues and fix a hermitian matrix $A\not\in S$. In the vector space of hermitian matrices, any line through $A$ intersects $S$ in at most finitely many points. From this it easily follows that $S$ is negligible (using polar coordinates centered at $A$).


To check the claim, note that a line through $A$ consists of matrices of the form $M(t)=(1-t)A+tB$, for some $B\in S$. Hence, the characteristic polynomial of $M(t)$ has the form $\lambda^n+\sum\_{k=0}^{n-1}p\_k(t)\lambda^k$ for some polynomials $p\_k(t)$.


There is a polynomial expression of the coefficients $a\_k$ of a polynomial $\lambda^n+\sum\_{k=0}^{n-1}a\_k\lambda^k$ which vanishes precisely when it has repeated roots: indeed, calling $\alpha\_i$ the roots, you can consider the expression $\prod\_{i<j}(\alpha\_i-\alpha\_j)^2$. Since it's symmetric, it is in fact equal to some $P(a\_0,\dots,a\_{n-1})$. For our matrices, this means that $M(t)\in S$ precisely when $P(p\_0(t),\dots,p\_{n-1}(t))=0$. The left-hand side is a polynomial, which is not trivial since it does not vanish at $t=0$. Hence, $M(t)\in S$ only for finitely many $t$.


