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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 $T : V \to V$ 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 $T : V \to V$ .

[L1]

Every self-adjoint endomorphism is normal (Self-adjoint and normal endomorphisms of a finite-dimensional real or complex inner product space).

[L2]

A real normal endomorphism admits an orthonormal block decomposition by $1 \times 1$ real blocks and $2 \times 2$ blocks $(\begin{array}{cc} a & b \\ - b & a \end{array})$ (A real normal endomorphism is orthogonally block-diagonalisable with 1x1 real blocks and 2x2 rotation-scaling blocks).

[L3]

In an orthonormal basis, self-adjointness means symmetry of the matrix (In an orthonormal basis, self-adjoint means conjugate-transpose symmetry and normal means commuting with the conjugate transpose).

Proof

technique · direct

1.1L1L2

By [L1] and [L2], there is an orthonormal basis in which the matrix of $T$ is block diagonal with $1 \times 1$ real blocks and $2 \times 2$ blocks $B = (\begin{array}{cc} a & b \\ - b & a \end{array})$ .

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 $2 \times 2$ block is the diagonal block $a I_{2} = diag (a, a)$ . 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

Dependency tree · 2 levels

Sources

Sheldon Axler, Linear Algebra Done Right, fourth edition