The Schur Decomposition: Every Matrix Is Unitarily Triangular

The precise statement

Schur's theorem (1909). Let A ∈ ℂⁿˣⁿ. Then there exist a unitary matrix U ∈ ℂⁿˣⁿ (meaning U*U = UU* = I, where U* is the conjugate transpose) and an upper-triangular matrix T ∈ ℂⁿˣⁿ such that

A = U T U*.

The pair (U, T) is a Schur decomposition of A, and T is a Schur form. Two facts pin down what T contains:

• Because U is unitary, A and T are unitarily similar, so they share the same characteristic polynomial. Hence the diagonal entries T₁₁, …, Tₙₙ are precisely the eigenvalues λ₁, …, λₙ of A, listed with algebraic multiplicity.
• The order of those eigenvalues along the diagonal is free: for any prescribed ordering of the spectrum there is a Schur decomposition realizing it. This orderability is what makes the theorem so useful in practice.

The decomposition is not unique — replacing U by U·diag(e^{iθ_k}) leaves a valid Schur form — but the diagonal spectrum is invariant.

The picture: rotate until it's triangular

Diagonalizing A means finding a basis of eigenvectors. The trouble is that a defective matrix simply doesn't have enough eigenvectors, and even when it does, an eigenbasis need not be orthogonal — so P⁻¹ can be wildly ill-conditioned. Schur's theorem asks for less and gets a guarantee in return: instead of demanding a full eigenbasis, we only build an orthonormal basis whose first k vectors span an A-invariant subspace for each k.

Geometrically, T being upper triangular says exactly this: the flag of subspaces V₁ ⊂ V₂ ⊂ ⋯ ⊂ Vₙ = ℂⁿ spanned by the columns of U consecutively is a complete flag of A-invariant subspaces. A(Vₖ) ⊆ Vₖ for every k. Working in an orthonormal basis adapted to this flag, A can only 'point forward or sideways,' never backward — the strictly-lower entries vanish. You are rotating your coordinate frame (never stretching it) until A becomes triangular.

The mechanism: one eigenvector, then deflate

The proof is a clean induction on n, and its engine is the existence of a single eigenvector — which is where ℂ being algebraically closed is used.

• Base case. n = 1: every 1×1 matrix is already triangular.
• One eigenvector. Over ℂ the characteristic polynomial det(λI − A) has a root λ₁, so there is a unit vector u₁ with A u₁ = λ₁ u₁.
• Extend to an orthonormal basis. By Gram–Schmidt, complete u₁ to an orthonormal basis; let U₁ = [u₁ | ⋯] be the unitary matrix with these columns. Then, because the first column is an eigenvector,

U₁* A U₁ = [ λ₁ w* ; 0 A′ ],

a block with a zero first column below λ₁ and an (n−1)×(n−1) block A′ (this is deflation).

• Induct. Apply the theorem to A′: A′ = U′ T′ U′*. Assemble Û = diag(1, U′). Then U = U₁ Û is unitary and U* A U is upper triangular.

Every step uses only Gram–Schmidt and one root — no assumption that A is diagonalizable.

A worked example: a defective 2×2

Take the defective matrix

A = [ 2 1 ; −1 0 ].

Its characteristic polynomial is λ² − 2λ + 1 = (λ − 1)², a repeated eigenvalue λ = 1. Solving (A − I)v = 0 gives (up to scale) only v = (1, −1)ᵀ — a single eigenvector, so A is not diagonalizable. Diagonalization is impossible; Schur still works.

Normalize the eigenvector: u₁ = (1/√2)(1, −1)ᵀ. Complete to an orthonormal basis with u₂ = (1/√2)(1, 1)ᵀ, and set U = (1/√2) [ 1 1 ; −1 1 ]. Then

T = U* A U = [ 1 2 ; 0 1 ].

This is upper triangular with the eigenvalue 1 twice on the diagonal, and the superdiagonal entry 2 records the 'defect' — the amount by which A fails to be diagonal. You can verify U T U* = A directly. Note T here is exactly a scaled Jordan block, but obtained by a unitary change of basis.

Why the hypotheses matter, and the link to the spectral theorem

Algebraic closure is essential. The proof needs one eigenvector, which needs a root of the characteristic polynomial. Over ℝ this can fail: the rotation R = [ 0 −1 ; 1 0 ] has eigenvalues ±i and no real eigenvector, so there is no real orthogonal Q with QᵀRQ upper triangular. The fix is the real Schur form: A = Q S Qᵀ with Q orthogonal and S block-upper-triangular, using 2×2 blocks for conjugate complex eigenvalue pairs.

The spectral theorem is a corollary. Suppose A is normal (A A* = A* A). Its Schur form T is also normal, and a normal upper-triangular matrix must be diagonal (compare diagonal entries of T T* and T* T row by row: the off-diagonal entries are forced to zero). So A = U D U* with D diagonal — A is unitarily diagonalizable. Thus the spectral theorem for Hermitian, unitary, and normal matrices falls straight out of Schur triangularization. Schur also cleanly proves that eigenvalues depend continuously on entries and yields the trace = ∑λᵢ, det = ∏λᵢ identities at a glance.

Why it matters: the workhorse of numerical linear algebra

Schur form is what your computer actually computes when you ask for eigenvalues. The industry-standard QR algorithm (Francis, Kublanovskaya, ~1961) iteratively drives a matrix toward its Schur form using only orthogonal (unitary) transformations — a numerically stable process precisely because unitary maps preserve the 2-norm and don't amplify rounding error, unlike the potentially ill-conditioned P in a diagonalization. LAPACK's core eigensolvers return exactly a Schur decomposition.

• Matrix functions. The Schur–Parlett algorithm evaluates f(A) = U f(T) U* for analytic f (matrix exponential, logarithm, square root) by working on the triangular T, where recurrences run along diagonals.
• Control theory. Solving Sylvester and algebraic Riccati equations, and the ordered Schur form (invariant-subspace deflation), underpin stabilization and the Kalman filter.
• Structured problems. The QZ / generalized Schur form handles the pencil A − λB for generalized eigenvalue problems.

Whenever eigenvalues appear in scientific computing, Schur — not Jordan, not diagonalization — is doing the real work.

Schur decomposition versus its neighbors: what each factorization guarantees and what it costs.

Factorization	Form	Always exists over ℂ?	Basis is orthonormal (unitary)?
Schur	A = U T U*, T upper triangular	Yes, for every square A	Yes
Diagonalization	A = P D P⁻¹, D diagonal	No — fails for defective A (e.g. Jordan block)	Only if A is normal
Spectral theorem	A = U D U*, D diagonal	Only if A is normal (A A* = A* A)	Yes
Jordan form	A = P J P⁻¹, J Jordan blocks	Yes over ℂ	No — P generally ill-conditioned
Real Schur	A = Q S Qᵀ, S block-upper-triangular	Yes for real A, staying real	Yes (Q orthogonal)