Schur disassembly

In linear algebra , a branch of mathematics , the Schur decomposition or Schur normal form (according to Issai Schur ) is an important matrix decomposition, more precisely a trigonalization method .

definition

${\displaystyle A}$be a square matrix with entries (i.e. , where either stands for or for ). If the characteristic polynomial of over breaks down into linear factors , there is a unitary matrix , so that ${\displaystyle \mathbb {K} }$ ${\displaystyle A\in \mathbb {K} ^{n\times n}}$ ${\displaystyle \mathbb {K} }$ ${\displaystyle \mathbb {R} }$ ${\displaystyle \mathbb {C} }$ ${\displaystyle A}$ ${\displaystyle \mathbb {K} }$ ${\displaystyle U\in \mathbb {K} ^{n\times n}}$

${\displaystyle R=U^{*}AU\quad }$( is the matrix to be adjoint ) ${\displaystyle U^{*}}$ ${\displaystyle U}$

is an upper triangular matrix . Since is unitary, it follows ; such a representation is called the Schur decomposition of . ${\displaystyle U}$ ${\displaystyle A=URU^{*}}$ ${\displaystyle A}$

Remarks

• Since is an upper triangular matrix, it can be represented as the sum of a diagonal matrix and a strict upper triangular matrix ( ): ${\displaystyle R}$ ${\displaystyle D}$ ${\displaystyle N}$ ${\displaystyle D,N\in \mathbb {K} ^{n\times n}}$

${\displaystyle R=D+N}$

The following then applies:

• ${\displaystyle D}$is unique except for the order of the diagonal elements and is referred to as the diagonal part of the Schur decomposition .
• ${\displaystyle N}$is nilpotent , generally only unique with regard to its Frobenius norm, and is called the nilpotent part of the Schur decomposition .
• The Frobenius norm of is 0 if and only if is normal . ${\displaystyle N}$ ${\displaystyle A}$

• Because of the similarity of the output matrix and the upper triangular matrix are on the main diagonal of the eigenvalues of . ${\displaystyle A}$ ${\displaystyle R}$ ${\displaystyle R}$ ${\displaystyle A}$
• If is a normal matrix, then is even a diagonal matrix and the column vectors of are eigenvectors of . The Schur decomposition of is then used as spectral decomposition of designated. ${\displaystyle A}$ ${\displaystyle R}$ ${\displaystyle U}$ ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle A}$
• If positive is definite , then the Schur decomposition of is the same as the singular value decomposition of . ${\displaystyle A}$ ${\displaystyle A}$ ${\displaystyle A}$

Construction of a Schur dismantling

Be . First, an eigenvalue and a corresponding eigenvector must be found. Vectors are now chosen so that they form an orthonormal basis in . These vectors form the columns of a matrix with ${\displaystyle A\in \mathbb {K} ^{n\times n}}$ ${\displaystyle \lambda _{1}}$ ${\displaystyle v_{1}}$ ${\displaystyle A}$ ${\displaystyle n-1}$ ${\displaystyle w_{2},\ldots ,w_{n}}$ ${\displaystyle v_{1},w_{2},\ldots ,w_{n}}$ ${\displaystyle \mathbb {K} ^{n}}$ ${\displaystyle V_{1}}$

${\displaystyle V_{1}^{*}AV_{1}={\begin{bmatrix}\lambda _{1}&*\\0&A_{1}\end{bmatrix}}}$,

where is a matrix. Now this process is repeated for. A unitary matrix is ​​created with ${\displaystyle A_{1}}$ ${\displaystyle (n-1)\times (n-1)}$ ${\displaystyle A_{1}}$ ${\displaystyle V_{2}}$

${\displaystyle V_{2}^{*}A_{1}V_{2}={\begin{bmatrix}\lambda _{2}&*\\0&A_{2}\end{bmatrix}}}$,

where is a matrix. Then applies ${\displaystyle A_{2}}$ ${\displaystyle (n-2)\times (n-2)}$

${\displaystyle Q_{2}^{*}AQ_{2}={\begin{bmatrix}\lambda _{1}&*&*\\0&\lambda _{2}&*\\0&0&A_{2}\end{bmatrix}}}$,

where with applies. The entire procedure is repeated times until the matrices are available. Then is a unitary matrix and an upper triangular matrix . This determines the Schur decomposition of the matrix . ${\displaystyle Q_{2}=V_{1}{\hat {V}}_{2}}$ ${\displaystyle {\hat {V}}_{2}={\begin{bmatrix}1&0\\0&V_{2}\end{bmatrix}}}$ ${\displaystyle (n-1)}$ ${\displaystyle V_{1},\ldots ,{\hat {V}}_{n-1}}$ ${\displaystyle Q:=V_{1}{\hat {V}}_{2}{\hat {V}}_{3}\cdots {\hat {V}}_{n-1}}$ ${\displaystyle R:=Q^{*}AQ}$ ${\displaystyle A}$

example

For example, consider the matrix with the eigenvalues (the matrix cannot be diagonalized because the dimension of the eigenspace associated with this eigenvalue is 1). ${\displaystyle A={\begin{bmatrix}-2&1&3\\2&1&-1\\-7&2&7\end{bmatrix}}}$ ${\displaystyle \lambda _{1}=\lambda _{2}=\lambda _{3}=2}$

We choose as the basis for the beginning of the standard base , where the designated th unit vector. ${\displaystyle \langle e_{1},e_{2},e_{3}\rangle }$ ${\displaystyle e_{j}}$ ${\displaystyle j}$

For we determine an eigenvector of 2, for example with representation, and add it to a linearly independent basis, e.g. B. . From this new basis we generate the basis transformation and from this it can be calculated that . ${\displaystyle A_{1}=A}$ ${\displaystyle {\begin{bmatrix}1\\1\\1\end{bmatrix}}}$ ${\displaystyle v_{1}:=1e_{1}+1e_{2}+1e_{3}={\begin{bmatrix}1\\1\\1\end{bmatrix}}}$ ${\displaystyle \langle v_{1},e_{1},e_{3}\rangle }$ ${\displaystyle V_{1}=(v_{1}|e_{1}|e_{3})}$ ${\displaystyle V_{1}^{-1}AV_{1}={\begin{bmatrix}2&2&-1\\0&-4&4\\0&-9&8\end{bmatrix}}}$ ${\displaystyle A_{2}={\begin{bmatrix}-4&4\\-9&8\end{bmatrix}}}$

For we determine an eigenvector of 2, e.g. B. with representation and supplement it to a linearly independent basis, z. B. . From this new basis we generate the basis transformation and calculate . ${\displaystyle A_{2}}$ ${\displaystyle {\begin{bmatrix}2\\3\end{bmatrix}}}$ ${\displaystyle v_{2}:=0v_{1}+2e_{1}+3e_{3}={\begin{bmatrix}2\\0\\3\end{bmatrix}}}$ ${\displaystyle \langle v_{1},v_{2},e_{3}\rangle }$ ${\displaystyle V_{2}=(v_{1}|v_{2}|e_{3})}$ ${\displaystyle V_{2}^{-1}AV_{2}={\begin{bmatrix}2&1&-1\\0&2&2\\0&0&2\end{bmatrix}}}$

As shown above, the base can be chosen arbitrarily, but things get very simple and interesting if the choice of the standard base is followed (if possible). This will change the previous steps as follows:

For we determine an eigenvector of 2, e.g. B. with representation and supplement it to a linearly independent basis, z. B. . From this new basis we generate the basis transformation and from this it can be calculated that . ${\displaystyle A_{1}=A}$ ${\displaystyle {\begin{bmatrix}1\\1\\1\end{bmatrix}}}$ ${\displaystyle v_{1}:={\begin{bmatrix}1\\1\\1\end{bmatrix}}}$ ${\displaystyle \langle v_{1},e_{2},e_{3}\rangle }$ ${\displaystyle V_{1}=(v_{1}|e_{2}|e_{3})}$ ${\displaystyle V_{1}^{-1}AV_{1}={\begin{bmatrix}2&1&3\\0&0&-4\\0&1&4\end{bmatrix}}}$ ${\displaystyle A_{2}={\begin{bmatrix}0&-4\\1&4\end{bmatrix}}}$

For we determine an eigenvector of 2, e.g. B. with representation and supplement it to a linearly independent basis, z. B. . From this new basis we generate the basis transformation and calculate . ${\displaystyle A_{2}}$ ${\displaystyle {\begin{bmatrix}2\\-1\end{bmatrix}}}$ ${\displaystyle v_{2}:={\begin{bmatrix}0\\2\\-1\end{bmatrix}}}$ ${\displaystyle \langle v_{1},v_{2},e_{3}\rangle }$ ${\displaystyle V_{2}=(v_{1}|v_{2}|e_{3})}$ ${\displaystyle V_{2}^{-1}AV_{2}={\begin{bmatrix}2&-1&3\\0&2&-2\\0&0&2\end{bmatrix}}}$

Here, the calculation of the representation of the vectors in the correct basis is, so to speak, intuitive and therefore less error-prone, and the final basis transformation is also a triangular matrix. ${\displaystyle V_{2}}$

With the Gram-Schmidt orthogonalization method, the obtained base transformation matrix can be made a unitary matrix as required.

Web links

• Eric W. Weisstein : Schur Decomposition . In: MathWorld (English).
• LP - Lemma from Schur , u. a. Proof of the lemma, in: Numerical Mathematics I - Functional Analysis Basics.