Symmetric matrices with distinct eigenvalues are open and dense
Statement
Show that the symmetric matrices with distinct eigenvalues form an open dense subset of .
Hint
Open: the discriminant is non-zero. Dense: diagonalise orthogonally, and perturb the eigenvalues in the diagonal factor.
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Solution
Idea. The spectral theorem lets us change the eigenvalues of a symmetric matrix at will without leaving : we keep the orthogonal factor and move the diagonal one.
Let .
Step 1: is open in . where is a polynomial in the entries (previous problems). It is the preimage of under a continuous map restricted to , hence open in .
Step 2: is dense in . Let . By the spectral theorem with orthogonal and . For put
This is symmetric (it is ), and its eigenvalues are . These are pairwise distinct unless for some , i.e. : finitely many values of to avoid. So for all small , . Finally, orthogonal invariance of the operator norm gives
Remark.
- Both statements together say that *a generic symmetric matrix has simple spectrum*, which is why proofs for symmetric matrices can usually assume distinct eigenvalues and conclude by continuity.
- The complement of in is closed with empty interior, but unlike in it has codimension , not : two eigenvalues of a symmetric matrix collide in codimension (von Neumann-Wigner "level repulsion"), e.g. in the repeated-eigenvalue matrices are the *line* of scalar matrices .