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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-28
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Real spectral theorem: a self-adjoint endomorphism of a finite-dimensional real inner product space has an orthonormal eigenbasis

Statement

Let V be a finite-dimensional real inner product space and let T:VV be self-adjoint. Then V has an orthonormal basis consisting of eigenvectors of T.

Facts & Assumptions

Given: A finite-dimensional real inner product space V and a self-adjoint endomorphism T:VV.

[L2]

A real normal endomorphism admits an orthonormal block decomposition by 1×1 real blocks and 2×2 blocks (abba) (A real normal endomorphism is orthogonally block-diagonalisable with 1x1 real blocks and 2x2 rotation-scaling blocks).

Proof

technique · direct
1.1

By [L1] and [L2], there is an orthonormal basis in which the matrix of T is block diagonal with 1×1 real blocks and 2×2 blocks B=(abba).

L1L2
2.1

Because T is self-adjoint, [L3] says that the same matrix is symmetric. A block B above is symmetric exactly when b=0, so every 2×2 block collapses to the scalar block [a]. Therefore the whole matrix is diagonal.

L3step 1.1algebra
3.1

A diagonal matrix acts on each basis vector by scalar multiplication, so the orthonormal basis from step 2.1 is an orthonormal eigenbasis of T.

step 2.1

Depends on

Used by

Dependency tree · two levels

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Sources