Sunday, September 24, 2023
HomeMatlabWhat Is the Schur Complement of a Matrix? – Nick Higham

What Is the Schur Complement of a Matrix? – Nick Higham

The lowered submatrices in Gaussian elimination are Schur enhances. Write an ntimes n matrix A with nonzero (1,1) aspect alpha as

notag   A =   begin{bmatrix} alpha & a^*                       b     & C   end{bmatrix}    =   begin{bmatrix}  1            & 0                          b/alpha     & I_{n-1}   end{bmatrix}   begin{bmatrix}  alpha       & a^*                         0           & C - ba^*/alpha   end{bmatrix}.

This factorization represents step one of Gaussian elimination.

After k phases of Gaussian elimination we’ve got computed the next factorization, wherein the (1,1) blocks are ktimes k:

notag   begin{bmatrix} L_{11} & 0                         L_{21} & I   end{bmatrix}   begin{bmatrix} A_{11} & A_{12}                    A_{21} & A_{22}   end{bmatrix}  =   begin{bmatrix} U_{11} & U_{12}                            0   & S   end{bmatrix},

the place the primary matrix on the left is the product of the elementary transformations that cut back the primary k columns to higher triangular type. Equating (2,1) and (2,2) blocks on this equation offers L_{21}A_{11} + A_{21} = 0 and L_{21}A_{12} + A_{22} = S. Therefore S= A_{22} + L_{21}A_{12}                       = A_{22} + (-A_{21}A_{11}^{-1})A_{12}, which is the Schur complement of A_{11} in A.

The following step of the elimination, which zeros out the primary column of S under the diagonal, succeeds so long as the (1,1) aspect of S is nonzero (or if the entire first column of S is zero, wherein case there’s nothing to do, and A is singular on this case).

For numerous matrix buildings, comparable to

  • Hermitian (or actual symmetric) constructive particular matrices,
  • completely constructive matrices,
  • matrices diagonally dominant by rows or columns,
  • M-matrices,

one can present that the Schur complement inherits the construction. For these 4 buildings the (1,1) aspect of the matrix is nonzero, so the success of Gaussian elimination (or equivalently, the existence of an LU factorization) is assured. Within the first three circumstances the preservation of construction can also be the idea for a proof of the numerical stability of Gaussian elimination.



Please enter your comment!
Please enter your name here

Most Popular

Recent Comments