Spectral Theorem

Statement

Let $A$ be a real symmetric $n \times n$ matrix (i.e., $A = A^T$). Then:

1. All eigenvalues of $A$ are real.
2. Eigenvectors corresponding to distinct eigenvalues are orthogonal.
3. There exists an orthogonal matrix $Q$ (so $Q^T = Q^{-1}$) and a diagonal matrix $\Lambda$ such that:

Equivalently, $A$ is orthogonally similar to a diagonal matrix of its eigenvalues. In the complex Hermitian setting ($A = A^*$), the same holds with a unitary matrix $U$ replacing $Q$.

Visualization

Take the symmetric matrix:

Step 1 — Find eigenvalues via $\det(\lambda I - A) = 0$:

Both eigenvalues are real. $\Lambda = \begin{pmatrix} 0 & 0 \\ 0 & 5 \end{pmatrix}$

Step 2 — Find eigenvectors.

For $\lambda_1 = 0$: $(A - 0)\mathbf{v} = 0$ gives $4v_1 + 2v_2 = 0$, so $\mathbf{v}_1 = \frac{1}{\sqrt{5}}\begin{pmatrix} -1 \\ 2 \end{pmatrix}$.

For $\lambda_2 = 5$: $(A - 5I)\mathbf{v} = 0$ gives $-v_1 + 2v_2 = 0$, so $\mathbf{v}_2 = \frac{1}{\sqrt{5}}\begin{pmatrix} 2 \\ 1 \end{pmatrix}$.

Note $\mathbf{v}_1 \cdot \mathbf{v}_2 = \frac{1}{5}(-2 + 2) = 0$: orthogonal, as guaranteed.

Step 3 — Assemble $Q$ and verify $A = Q\Lambda Q^T$.

Proof Sketch

The key steps for real symmetric matrices are: (1) every symmetric operator on a real finite-dimensional inner product space has at least one real eigenvalue (proved via the characteristic polynomial having a real root, using the fundamental theorem of algebra over $\mathbb{C}$ then checking reality). (2) By induction on dimension — restrict to the orthogonal complement of one eigenvector; the restricted operator is still symmetric. (3) Orthogonality of eigenvectors for distinct eigenvalues follows from $\lambda_1 \langle u, v \rangle = \langle Au, v \rangle = \langle u, Av \rangle = \lambda_2 \langle u, v \rangle$.

Connections

• Cayley–Hamilton TheoremCayley–Hamilton Theorem$$p_A(A) = 0 \;\text{where}\; p_A(\lambda) = \det(\lambda I - A)$$Every square matrix satisfies its own characteristic polynomialRead more → — the characteristic polynomial whose roots are these eigenvalues is also annihilated by $A$.
• Determinant MultiplicativityDeterminant Multiplicativity$$\det(AB) = \det(A)\,\det(B)$$The determinant of a product equals the product of determinantsRead more → — $\det(A) = \prod_i \lambda_i$; the product of diagonal entries of $\Lambda$.
• Rank–Nullity TheoremRank–Nullity Theorem$$\operatorname{rank}(f) + \operatorname{nullity}(f) = \dim V$$The dimensions of image and kernel of a linear map sum to the domain dimensionRead more → — the number of zero eigenvalues equals the nullity of $A$.
• Fundamental Theorem of Galois TheoryFundamental Theorem of Galois Theory$$\{\text{intermediate fields}\} \longleftrightarrow \{\text{subgroups of } \text{Gal}(K/F)\}$$Bijection between intermediate fields and subgroups of the Galois groupRead more → — the eigenvalue decomposition is the linear algebra analogue of a splitting field decomposition.
• Spectral Radius FormulaSpectral Radius Formula$$r(a) = \lim_{n\to\infty} \|a^n\|^{1/n}$$The spectral radius equals the limit of the nth root of the operator norm of the nth power (Gelfand's formula)Read more → — for bounded operators on a Banach space, the spectral radius $r(T) = \lim_{n\to\infty}\|T^n\|^{1/n}$ equals the supremum of $|\lambda|$ over all $\lambda$ in the spectrum; for Hermitian matrices this reduces to $r(A) = \|A\|_\text{op} = \max_i |\lambda_i|$.

/-- The spectral theorem for Hermitian matrices: a Hermitian matrix A is
    unitarily conjugate to a diagonal matrix of its (real) eigenvalues.
    Mathlib packages this as `Matrix.IsHermitian.spectral_theorem` in
    Mathlib.Analysis.Matrix.Spectrum. -/
noncomputable example
    {n 𝕜 : Type*} [RCLike 𝕜] [Fintype n] [DecidableEq n]
    {A : Matrix n n 𝕜} (hA : A.IsHermitian) :
    A = hA.eigenvectorUnitary.val *
        Matrix.diagonal (RCLike.ofReal ∘ hA.eigenvalues) *
        (star hA.eigenvectorUnitary.val) :=
  hA.spectral_theorem