Schur decomposition

From Free net encyclopedia

(Redirected from Schur triangulation)

In the mathematical discipline of linear algebra, the Schur decomposition or Schur triangulation (named after Issai Schur) is an important matrix decomposition.

Contents

Definition

If A is a square matrix over the complex numbers, then A can be decomposed as

<math>A= QUQ^*, \,</math>

where Q is a unitary matrix, Q* is the conjugate transpose of Q and U is an upper triangular matrix whose diagonal entries are exactly the eigenvalues of A.

Notes

Every square matrix has a Schur decomposition, and hence, every square matrix is unitarily equivalent to a triangular matrix (indeed, Q*AQ = U). However, this decomposition is not unique.

Write the triangular matrix U as U = D + N, where D is diagonal and N is strictly upper triangular (and thus nilpotent). The diagonal matrix D contains the eigenvalues of A in arbitrary order. Furthermore, the nilpotent part N is generally not unique either, but its Frobenius norm is uniquely determined by A.

If A is a normal matrix, then U is even a diagonal matrix and the column vectors of Q are the eigenvectors of A and the Schur decomposition is called the spectral decomposition. Furthermore, if A is positive definite, the Schur decomposition of A is the same as the singular value decomposition of the matrix.

A commuting family of matrices can be simultaneously triangularized. This means that, given several commuting matrices A1, …, An, there exists a unitary matrix Q such that the matrices Q*A1Q, …, Q*AnQ are all upper triangular.

Construction of the Schur decomposition

Some algorithms in numerical linear algebra require a means of computing a Schur decomposition for a matrix. This can be done by the following procedure, which also shows that a Schur decomposition exists.

Given the n-by-n matrix A, find an eigenvalue λ1 of A with corresponding eigenvector v1 of unit norm. Choose n − 1 vectors w2, …, wn, such that

<math> v_1, w_2, w_3, \ldots, w_n \, </math>

is an orthonormal basis for Cn. If V1 denotes the matrix with these vectors as columns, then

<math> V_1^* A V_1 = \begin{bmatrix} \lambda_1 & * \\ 0 & A_1 \end{bmatrix} </math>

where <math>A_1</math> is an (n − 1)-by-(n − 1) matrix.

Now, repeat this process with A1: this gives a unitary matrix V2 such that

<math> V_2^* A_1 V_2 = \begin{bmatrix} \lambda_2 & * \\ 0 & A_2 \end{bmatrix} </math>

where <math>A_2</math> is an (n − 2)-by-(n − 2) matrix. Hence,

<math> Q_2^* A Q_2 = \begin{bmatrix} \lambda_1 & * & * \\ 0 & \lambda_2 & * \\ 0 & 0 & A_2 \end{bmatrix}, \quad\mbox{where } Q_2 = V_1 \hat{V}_2 \mbox{ with } \hat{V}_2 = \begin{bmatrix} 1 & 0 \\ 0 & V_2 \end{bmatrix}. </math>

Continuing this process, one finds the matrices V3, …, Vn. Finally, the matrix U = Q*AQ with

<math> Q = V_1 \hat{V}_2 \hat{V}_3 \cdots \hat{V}_n </math>

is upper triangular, so A = QUQ* is a Schur decomposition of A.

References

it:Decomposizione di Schur