# Schur decomposition

In the mathematical discipline of linear algebra, the Schur decomposition or Schur triangulation, named after Issai Schur, is a matrix decomposition. It allows one to write an arbitrary complex square matrix as unitarily similar to an upper triangular matrix whose diagonal elements are the eigenvalues of the original matrix.

## Statement

The complex Schur decomposition reads as follows: if A is an n × n square matrix with complex entries, then A can be expressed as $A=QUQ^{-1}$ for some unitary matrix Q (so that the inverse Q$^{−1}$ is also the conjugate transpose Q* of Q), and some upper triangular matrix U. This is called a Schur form of A. Since U is similar to A, it has the same spectrum, and since it is triangular, its eigenvalues are the diagonal entries of U.
The Schur decomposition implies that there exists a nested sequence of A-invariant subspaces {0} = V$_0$ ⊂ V$_1$ ⊂ ⋯ ⊂ V$_n$ = C$^n$, and that there exists an ordered orthonormal basis (for the standard Hermitian form of C$^n$) such that the first i basis vectors span V$_i$ for each i occurring in the nested sequence. Phrased somewhat differently, the first part says that a linear operator J on a complex finite-dimensional vector space stabilizes a complete flag (V$_1$, ..., V$_n$).
There is also a real Schur decomposition. If A is an n × n square matrix with real entries, then A can be expressed as $A=QHQ^{-1}$ where Q is an orthogonal matrix and H is either upper or lower quasi-triangular. A quasi-triangular matrix is a matrix that when expressed as a block matrix of 2 × 2 and 1 × 1 blocks is triangular. This is a stronger property than being Hessenberg . Just as in the complex case, a family of commuting real matrices {A$_i$} may be simultaneously brought to quasi-triangular form by an orthogonal matrix. There exists an orthogonal matrix Q such that, for every A$_i$ in the given family, $H_{i}=QA_{i}Q^{-1}$ is upper quasi-triangular.

## Proof

A constructive proof for the Schur decomposition is as follows: every operator A on a complex finite-dimensional vector space has an eigenvalue λ, corresponding to some eigenspace V$_λ$. Let V$_λ$$^⊥$ be its orthogonal complement. It is clear that, with respect to this orthogonal decomposition, A has matrix representation (one can pick here any orthonormal bases Z$_1$ and Z$_2$ spanning V$_λ$ and V$_λ$$^⊥$ respectively) ${\begin{bmatrix}Z_{1}&Z_{2}\end{bmatrix}}^{*}A{\begin{bmatrix}Z_{1}&Z_{2}\end{bmatrix}}={\begin{bmatrix}\lambda \,I_{\lambda }&A_{12}\\0&A_{22}\end{bmatrix}}:{\begin{matrix}V_{\lambda }\\\oplus \\V_{\lambda }^{\perp }\end{matrix}}\rightarrow {\begin{matrix}V_{\lambda }\\\oplus \\V_{\lambda }^{\perp }\end{matrix}}$ where I$_λ$ is the identity operator on V$_λ$. The above matrix would be upper-triangular except for the A$_{22}$ block. But exactly the same procedure can be applied to the sub-matrix A$_{22}$, viewed as an operator on V$_λ$$^⊥$, and its submatrices. Continue this way until the resulting matrix is upper triangular. Since each conjugation increases the dimension of the upper-triangular block by at least one, this process takes at most n steps. Thus the space C$^n$ will be exhausted and the procedure has yielded the desired result.
The above argument can be slightly restated as follows: let λ be an eigenvalue of A, corresponding to some eigenspace V$_λ$. A induces an operator T on the quotient space C$^n$/V$_λ$. This operator is precisely the A$_{22}$ submatrix from above. As before, T would have an eigenspace, say W$_μ$ ⊂ C$^n$ modulo V$_λ$. Notice the preimage of W$_μ$ under the quotient map is an invariant subspace of A that contains V$_λ$. Continue this way until the resulting quotient space has dimension 0. Then the successive preimages of the eigenspaces found at each step form a flag that A stabilizes.