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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The eigenvalues of the orthogonal compression of a self-adjoint endomorphism to a hyperplane interlace those of the original endomorphism

Statement

Let T:V→V be self-adjoint on an n-dimensional real inner product space, let W⊂V be a hyperplane, and let

C:=PW∘T∣W:W→W

be the orthogonal compression, where PW is the orthogonal projection onto W. If the eigenvalues of T and C are ordered by

λ1≥⋯≥λn,μ1≥⋯≥μn−1,

then

λi≥μi≥λi+1(1≤i≤n−1).

Facts & Assumptions

Given: A self-adjoint endomorphism T:V→V on an n-dimensional real inner product space, a hyperplane W⊂V, and its orthogonal projection PW.

[L2]

Courant-Fischer characterises the ordered eigenvalues of a self-adjoint operator by min-max formulas (Courant-Fischer min-max principle for self-adjoint endomorphisms on finite-dimensional real inner product spaces).

Proof

technique · direct
1.1L1algebra

For u,v∈W, the decomposition in [L1] gives ⟨PWTu,v⟩=⟨Tu,v⟩=⟨u,Tv⟩=⟨u,PWTv⟩, so C is self-adjoint on W. Also RC(u)=⟨Cu,u⟩⟨u,u⟩=⟨Tu,u⟩⟨u,u⟩=RT(u) for every nonzero u∈W.

2.1L2step 1.1∎

For any 1≤i≤n−1, Courant-Fischer in W and step 1.1 give μi=max⁡dim⁡U=i, U⊆Wmin⁡0≠u∈URC(u)=max⁡dim⁡U=i, U⊆Wmin⁡0≠u∈URT(u)≤λi, because the maximisation is over fewer i-dimensional subspaces than in [L2]. Likewise μi=min⁡dim⁡L=n−i, L⊆Wmax⁡0≠u∈LRC(u)=min⁡dim⁡L=n−i, L⊆Wmax⁡0≠u∈LRT(u)≥λi+1, because n−i=(n−1)−i+1 is the subspace dimension appearing in Courant-Fischer for the (i+1)-st eigenvalue of T. Hence λi≥μi≥λi+1.

Depends on

Used by

Dependency tree · two levels

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Sources