Schur decomposition

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

1 Definition
2 Notes
3 Construction of the Schur decomposition
4 References

[edit] Definition

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

$A = Q U Q^*\,$

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.

[edit] 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, …, A_n, there exists a unitary matrix Q such that the matrices Q^*A₁Q, …, Q^*A_nQ are all upper triangular.

[edit] 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, …, w_n, such that

$v_1, w_2, w_3, \ldots, w_n \,$

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

$V_1^* A V_1 = \begin{bmatrix} \lambda_1 & * \\ 0 & A_1 \end{bmatrix}$

where $A 1$ is an (n − 1)-by-(n − 1) matrix.

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

$V_2^* A_1 V_2 = \begin{bmatrix} \lambda_2 & * \\ 0 & A_2 \end{bmatrix}$

where $A 2$ is an (n − 2)-by-(n − 2) matrix. Hence,

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