Schur Decomposition — Mathematics Knowledge System

Overview

The Schur decomposition states that every square complex matrix A can be written as A = Q T Q* where Q is unitary and T is upper triangular with the eigenvalues of A on the diagonal. Unlike diagonalization, Schur decomposition always exists for any square matrix. For normal matrices, T is diagonal — recovering the spectral theorem. The QR algorithm computes the Schur form numerically.

Intuition

Diagonalization is ideal but not always possible. Schur decomposition is the next best thing: every matrix can be brought to upper triangular form by a unitary (length-preserving) change of basis. The eigenvalues appear on the diagonal of T in any order. The off-diagonal entries of T are the 'mixing' that remains when eigenvectors are not orthogonal. For normal matrices, that mixing vanishes.

Formal Definition

Definition

For any A ∈ M_n(C), there exists a unitary Q and upper triangular T such that:

A = Q T Q^*, \quad Q \text{ unitary}, \quad T \text{ upper triangular}

Schur decomposition

t_{ii} = \lambda_i \quad (\text{eigenvalues of } A \text{ on diagonal of } T)

Eigenvalues on diagonal

\text{If } A \text{ is normal: } T = D \text{ (diagonal)} \iff A = QDQ^* \text{ (unitary diagonalization)}

Normal case

Notation

Notation	Meaning
$Q$	Unitary matrix of Schur vectors (columns form orthonormal basis)
$T$	Upper triangular Schur form; eigenvalues on diagonal

Properties

Always exists

\text{Unlike diagonalization, Schur decomposition exists for every square complex matrix.}

Eigenvalues preserved

\text{The diagonal of } T \text{ contains exactly the eigenvalues of } A \text{ (with multiplicity).}

Not unique

\text{The eigenvalue ordering on the diagonal can be chosen, and } Q \text{ is not unique.}

Real Schur form

\text{Real matrices have a real Schur form: } A = Q R Q^\top \text{ with } Q \text{ orthogonal and } R \text{ block upper triangular (2×2 real blocks for complex eigenvalue pairs).}

Theorems

Theorem 1: Schur's Theorem

\text{Every } A \in M_n(\mathbb{C}) \text{ has a Schur decomposition } A = QTQ^* \text{ with } Q \text{ unitary and } T \text{ upper triangular.}

Worked Examples

1
A is already upper triangular, so T = A and Q = I (the identity) is trivially unitary.
$A = I \cdot A \cdot I^* = \begin{pmatrix}3&1\\0&2\end{pmatrix}$
2
Eigenvalues are the diagonal entries: λ_1=3, λ_2=2.

✓ Answer

Q = I, T = A = [[3,1],[0,2]]; eigenvalues 3 and 2 on diagonal.

Practice Problems

Mediumfree response

The matrix A = [[4,2],[0,4]] is already in upper triangular form. Identify its Schur decomposition, its eigenvalues, and state whether A is diagonalizable.

Mediumproof writing

Using the Schur decomposition A = QTQ*, show that det(A) = product of diagonal entries of T.

Common Mistakes

Common Mistake

Schur decomposition means the matrix is diagonalizable.

Schur form T is upper triangular, not necessarily diagonal. Diagonalization requires T to be diagonal, which happens only for normal (or diagonalizable) matrices.

Quiz

Summary

Schur decomposition: every complex matrix A = QTQ* with Q unitary and T upper triangular.
Eigenvalues of A appear on the diagonal of T.
Schur decomposition always exists; diagonalization does not.
For normal matrices (AA* = A*A), T is diagonal — this is the spectral theorem.
Real matrices have a real Schur form with orthogonal Q and block upper triangular T.

References

WebsiteWikipedia — Schur decomposition