TheoremProved

Spectral theorem (real symmetric matrices)

Statement

Every real symmetric matrix $A$ (i.e. $A^{\mathsf T}=A$ ) is orthogonally diagonalizable: there is an orthogonal matrix $Q$ and a diagonal matrix $D$ with real entries such that $A = QDQ^{\mathsf T}$ ; equivalently, $A$ has an orthonormal basis of real eigenvectors.

Why is it true?

A symmetric matrix never twists space in the way a general matrix can — it only stretches along a set of mutually perpendicular axes. The spectral theorem says those axes always exist and are enough to describe the whole action of $A$ : rotate to align with them (the orthogonal $Q$ ), stretch each one by its own real factor (the diagonal $D$ ), then rotate back.

Proof sketch

Induction on dimension $n$ : since $A$ is real symmetric, its characteristic polynomial has a real root $\lambda_1$ (a short argument with Hermitian inner products rules out non-real eigenvalues), giving a unit eigenvector $v_1$ . The orthogonal complement of $v_1$ is invariant under $A$ (because $A$ is symmetric), so restrict $A$ to that $(n-1)$ -dimensional subspace and apply the inductive hypothesis to build the remaining orthonormal eigenvectors.

Proved by

proved 1829Augustin-Louis Cauchy

Topics that use this theorem

Eigenvalues and diagonalization

Related theorems

Cayley–Hamilton theorem

Step-by-step proofs

No step-by-step proof yet for this theorem.

References

Roger A. Horn, Charles R. Johnson (2012). Matrix Analysis · DOI:10.1017/CBO9781139020411