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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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Weyl inequalities bound the eigenvalues of a sum of self-adjoint endomorphisms

Statement

Let A,B be self-adjoint endomorphisms of an n-dimensional real inner product space. Order the eigenvalues of A, B, and A+B by

α1αn,β1βn,γ1γn.

Then, for integers 1i,jn:

  1. If i+j1n, then γi+j1αi+βj.
  2. If i+jn1, then γi+jnαi+βj.

Facts & Assumptions

Given: Self-adjoint endomorphisms A and B of an n-dimensional real inner product space.

[L1]

Proof

technique · direct
1.1

Apply [L1] separately to A and B. There are subspaces Ui,Vj of dimensions ni+1,nj+1 such that every nonzero vector in Ui has Rayleigh quotient for A at most αi and every nonzero vector in Vj has Rayleigh quotient for B at most βj; likewise there are subspaces Ui+,Vj+ of dimensions i,j such that every nonzero vector in them has Rayleigh quotient at least αi,βj respectively.

L1
1.2

If i+j1n, then [L2] gives dim(UiVj)(ni+1)+(nj+1)n=nij+2=n(i+j1)+1. Every nonzero x in that intersection satisfies RA(x)αi and RB(x)βj, hence RA+B(x)=RA(x)+RB(x)αi+βj. Therefore γi+j1αi+βj by the min-over-subspaces form of [L1] for A+B.

L1L2algebra
2.1

If i+jn1, then [L2] gives dim(Ui+Vj+)i+jn. Every nonzero x in that intersection satisfies RA(x)αi and RB(x)βj, hence RA+B(x)=RA(x)+RB(x)αi+βj. Therefore γi+jnαi+βj by the max-over-subspaces form of [L1] for A+B.

L1L2algebra

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