Schur Decomposition

Statement

For every $A \in \mathbb{C}^{n \times n}$ there exists a unitary matrix $Q$ (i.e.\ $Q Q^* = I$) and an upper triangular matrix $T$ such that

The diagonal entries of $T$ are precisely the eigenvalues of $A$ (with multiplicity). When $A$ is normal ($A A^* = A^* A$) the triangular form $T$ is actually diagonal, recovering the Spectral TheoremSpectral Theorem$$A = Q \Lambda Q^T \;\text{(real symmetric case)}$$Every real symmetric matrix is orthogonally diagonalizableRead more →.

Visualization

Take $A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$, the standard rotation by $90°$. Over $\mathbb{R}$ it has no eigenvalues, but over $\mathbb{C}$ it has eigenvalues $\pm i$.

Eigenvalues of A: λ₁ = i,  λ₂ = -i
Eigenvectors (columns of Q):
  q₁ = (1/√2)(1, -i)ᵀ,   q₂ = (1/√2)(1, i)ᵀ

Schur form T = QᴴAQ:
  T = ( i    *  )     (* = off-diagonal entry, value 0 here since A is normal)
      ( 0   -i  )

Because $A$ is unitary (hence normal), the Schur form is diagonal:

Step	Value
$A q_1$	$i \, q_1$ (eigenvector)
$A q_2$	$-i \, q_2$ (eigenvector)
$Q^* A Q$	$\text{diag}(i, -i) = T$

Proof Sketch

1. Base case. Pick any eigenvalue $\lambda_1$ and unit eigenvector $q_1$. Extend to an orthonormal basis $\{q_1, \ldots, q_n\}$ of $\mathbb{C}^n$.
2. Block reduction. Relative to this basis, $A$ has the block form $\begin{pmatrix} \lambda_1 & * \\ 0 & A' \end{pmatrix}$ where $A'$ is $(n-1)\times(n-1)$.
3. Induction. Apply the construction to $A'$. By induction, $A'$ is unitarily similar to an upper triangular matrix $T'$, which assembles into $T = \begin{pmatrix} \lambda_1 & * \\ 0 & T' \end{pmatrix}$.
4. Diagonal for normal matrices. If $A$ is normal, $\|T e_j\|^2 = \|T^* e_j\|^2$ for every standard basis vector forces all off-diagonal entries to zero, giving a diagonal $T$.

Connections

• Spectral TheoremSpectral Theorem$$A = Q \Lambda Q^T \;\text{(real symmetric case)}$$Every real symmetric matrix is orthogonally diagonalizableRead more → — Schur decomposition restricted to normal matrices is exactly the spectral theorem; the diagonal form records eigenvalues
• 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 of $A$ equals that of its Schur form $T$ (since $\det(A - \lambda I) = \det(T - \lambda I)$); Cayley–Hamilton then applies equally to $T$
• Jordan Canonical FormJordan Canonical Form$$A = P J P^{-1},\quad J = \bigoplus_i J_{n_i}(\lambda_i)$$Every square matrix over an algebraically closed field is similar to a direct sum of Jordan blocks.Read more → — Schur triangularises without ordering eigenspaces; Jordan goes further, revealing the nilpotent structure inside each eigenspace

import Mathlib.Analysis.Matrix.Spectrum

/-- Schur decomposition for Hermitian matrices (the normal case):
    the star-algebra automorphism by the conjugate of the eigenvector
    unitary diagonalizes A.  This is `conjStarAlgAut_star_eigenvectorUnitary`
    in Mathlib — a direct one-line alias. -/
theorem hermitian_schur
    {𝕜 : Type*} [RCLike 𝕜]
    {n : Type*} [Fintype n] [DecidableEq n]
    {A : Matrix n n 𝕜} (hA : A.IsHermitian) :
    Unitary.conjStarAlgAut 𝕜 _ (star hA.eigenvectorUnitary) A =
      Matrix.diagonal (RCLike.ofReal ∘ hA.eigenvalues) :=
  hA.conjStarAlgAut_star_eigenvectorUnitary