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Exercise 2060Topology › Topology of Mn(ℝ) › ConnectednessL3Added 7 Nov 2026

Components of the invertible symmetric matrices

#symmetric matrices#signature#eigenvalues#Weyl inequality

Statement

Show that the set of invertible symmetric matrices in Mn(R)M_n(\mathbb{R}) has exactly n+1n+1 path components, indexed by the number k∈{0,…,n}k\in\{0,\dots,n\} of positive eigenvalues.

Hint

Show that kk is locally constant, and that each class {k positive eigenvalues}\{k\text{ positive eigenvalues}\} is the image of a connected set under (Q,d)↦Qdiag⁡(d)QT(Q,d)\mapsto Q\operatorname{diag}(d)Q^T.

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Solution

Idea. The signature is a continuous integer-valued invariant (the eigenvalues cannot cross 00 without making the matrix singular), and matrices with the same signature are connected through the spectral theorem.

Step 1: continuity of eigenvalues. Let Symn∗\mathrm{Sym}_n^* be the set of invertible symmetric matrices and λ1(S)≥⋯≥λn(S)\lambda_1(S)\ge\dots\ge\lambda_n(S) the ordered eigenvalues of SS. By *Weyl's inequality*,
∣λi(S)−λi(S′)∣≤∥S−S′∥(operator norm),|\lambda_i(S)-\lambda_i(S')|\le\|S-S'\|\quad(\text{operator norm}),
so each λi\lambda_i is continuous (even 11-Lipschitz) on Symn\mathrm{Sym}_n.

Step 2: the classes are clopen. On Symn∗\mathrm{Sym}_n^* no λi\lambda_i vanishes. The number of positive eigenvalues is k(S)=#{i:λi(S)>0}k(S)=\#\{i:\lambda_i(S)>0\}. Since each λi\lambda_i is continuous and never 00 on Symn∗\mathrm{Sym}_n^*, the sign of each λi\lambda_i is locally constant, hence so is kk. So the classes Ck={S∈Symn∗:k(S)=k}\mathcal{C}_k=\{S\in\mathrm{Sym}_n^*:k(S)=k\}, k=0,…,nk=0,\dots,n, are open, disjoint, non-empty (they contain diag⁡(Ik,−In−k)\operatorname{diag}(I_k,-I_{n-k})) and cover Symn∗\mathrm{Sym}_n^*; each is also closed in Symn∗\mathrm{Sym}_n^* as a complement of the union of the others.

Step 3: each class is connected. By the spectral theorem every S∈CkS\in\mathcal{C}_k is S=Qdiag⁡(d)QTS=Q\operatorname{diag}(d)Q^T with Q∈O(n)Q\in O(n) and d∈Δk=(0,∞)k×(−∞,0)n−kd\in\Delta_k=(0,\infty)^k\times(-\infty,0)^{n-k}.
- *We may take Q∈SO(n)Q\in SO(n).* Let D′=diag⁡(−1,1,…,1)D'=\operatorname{diag}(-1,1,\dots,1). Since D′diag⁡(d)D′=diag⁡(d)D'\operatorname{diag}(d)D'=\operatorname{diag}(d), S=(QD′)diag⁡(d)(QD′)TS=(QD')\operatorname{diag}(d)(QD')^T and one of Q,QD′Q,QD' has determinant 11.
- *Connected parameter space.* SO(n)SO(n) is path connected (Problem 16) and Δk\Delta_k is convex. So SO(n)×ΔkSO(n)\times\Delta_k is path connected, and Ck\mathcal{C}_k is its continuous image under (Q,d)↦Qdiag⁡(d)QT(Q,d)\mapsto Q\operatorname{diag}(d)Q^T.

Conclusion. The path components are exactly C0,…,Cn\mathcal{C}_0,\dots,\mathcal{C}_n: n+1n+1 of them.

Remark. Cn\mathcal{C}_n is the cone of positive definite matrices and C0\mathcal{C}_0 its negative. Ck\mathcal{C}_k for 0<k<n0<k<n are the sets of *quadratic forms of signature (k,n−k)(k,n-k)*, in the classification of quadratic forms by Sylvester's law of inertia. For n=2n=2 there are three components: the two cones {ac>b2,a>0}\{ac>b^2,a>0\}, {ac>b2,a<0}\{ac>b^2,a<0\} and the "inside of the light cone" {ac<b2}\{ac<b^2\}.