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Exercise 2045Topology › Topology of Mn(ℝ) › Open, closed, interior, boundaryM1Added 7 Nov 2026

Symmetric matrices with distinct eigenvalues are open and dense

#symmetric matrices#spectral theorem#genericity#distinct eigenvalues

Statement

Show that the symmetric matrices with nn distinct eigenvalues form an open dense subset of Symn\mathrm{Sym}_n.

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 Symn\mathrm{Sym}_n: we keep the orthogonal factor and move the diagonal one.

Let G={S∈Symn:S has n distinct eigenvalues}\mathcal G=\{S\in\mathrm{Sym}_n:S\text{ has }n\text{ distinct eigenvalues}\}.

Step 1: G\mathcal G is open in Symn\mathrm{Sym}_n. G={S∈Symn:Δ(S)≠0}\mathcal G=\{S\in\mathrm{Sym}_n:\Delta(S)\ne0\} where Δ(S)=disc⁡(χS)\Delta(S)=\operatorname{disc}(\chi_S) is a polynomial in the entries (previous problems). It is the preimage of R∖{0}\mathbb{R}\setminus\{0\} under a continuous map restricted to Symn\mathrm{Sym}_n, hence open in Symn\mathrm{Sym}_n.

Step 2: G\mathcal G is dense in Symn\mathrm{Sym}_n. Let S∈SymnS\in\mathrm{Sym}_n. By the spectral theorem S=QΛQTS=Q\Lambda Q^T with QQ orthogonal and Λ=diag⁡(λ1,…,λn)\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_n). For ε≠0\varepsilon\ne0 put
Sε=Q(Λ+εdiag⁡(1,2,…,n))QT.S_\varepsilon=Q\big(\Lambda+\varepsilon\operatorname{diag}(1,2,\dots,n)\big)Q^T.
This is symmetric (it is Q diag QTQ\,\mathrm{diag}\,Q^T), and its eigenvalues are λi+iε\lambda_i+i\varepsilon. These are pairwise distinct unless λi+iε=λj+jε\lambda_i+i\varepsilon=\lambda_j+j\varepsilon for some i<ji<j, i.e. ε=λi−λjj−i\varepsilon=\frac{\lambda_i-\lambda_j}{j-i}: finitely many values of ε\varepsilon to avoid. So for all small ε≠0\varepsilon\ne0, Sε∈GS_\varepsilon\in\mathcal G. Finally, orthogonal invariance of the operator norm gives
∥Sε−S∥=ε ∥diag⁡(1,…,n)∥=nε→0.\|S_\varepsilon-S\|=\varepsilon\,\|\operatorname{diag}(1,\dots,n)\|=n\varepsilon\to0.

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 G\mathcal G in Symn\mathrm{Sym}_n is closed with empty interior, but unlike in Mn(R)M_n(\mathbb{R}) it has codimension 22, not 11: two eigenvalues of a symmetric matrix collide in codimension 22 (von Neumann-Wigner "level repulsion"), e.g. in Sym2≅R3\mathrm{Sym}_2\cong\mathbb{R}^3 the repeated-eigenvalue matrices are the *line* of scalar matrices {aI}\{aI\}.