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CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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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:V→V 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:V→V.

[L2]

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

Proof

technique · direct
1.1L1L2

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=(ab−ba).

2.1L3step 1.1algebra

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.

3.1step 2.1∎

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.

Depends on

Used by

Dependency tree · two levels

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Sources