A [[ B]. D M = where A E Rkxk is nonsingular. a) Verify that the following "block LU decomposition" formula holds: M = I A - [CA-+ i] [4 [CA-₁ в BA-'B]. D-CA-¹B The matrix D-CA-¹B is called the Schur complement of A and is the matrix we get after B eliminating the first block of unknowns x₁ in the system via the formula C D X₂ x₁ = A-¹(b₁ - Bx2); plugging this formula into the second block row of the equation yields: b₂ = Cx₁ + Dx2 = CA¯¹(b₁ – Bx2) + Dx2 (D- - CA¹B)x2 = b₂ - CA-¹b₁. b) Describe how you can construct the full LU decomposition of M by computing an LU decomposition of A, evaluating and factorizing D - CA-¹B, and finally manipulating the result using efficient operations (and never explicitly constructing A-¹)!
A [[ B]. D M = where A E Rkxk is nonsingular. a) Verify that the following "block LU decomposition" formula holds: M = I A - [CA-+ i] [4 [CA-₁ в BA-'B]. D-CA-¹B The matrix D-CA-¹B is called the Schur complement of A and is the matrix we get after B eliminating the first block of unknowns x₁ in the system via the formula C D X₂ x₁ = A-¹(b₁ - Bx2); plugging this formula into the second block row of the equation yields: b₂ = Cx₁ + Dx2 = CA¯¹(b₁ – Bx2) + Dx2 (D- - CA¹B)x2 = b₂ - CA-¹b₁. b) Describe how you can construct the full LU decomposition of M by computing an LU decomposition of A, evaluating and factorizing D - CA-¹B, and finally manipulating the result using efficient operations (and never explicitly constructing A-¹)!
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
we have a matrix M ∈ R ^n×n decomposed into blocks:
![Given the matrix \( M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \), where \( A \in \mathbb{R}^{k \times k} \) is nonsingular, we proceed with the following problems:
**a) Verify the "block LU decomposition" formula:**
The block LU decomposition is given by:
\[
M = \begin{bmatrix} I & 0 \\ CA^{-1} & I \end{bmatrix} \begin{bmatrix} A & B \\ 0 & D - CA^{-1}B \end{bmatrix}.
\]
The matrix \( D - CA^{-1}B \) is referred to as the **Schur complement** of \( A \) and is obtained after eliminating the first block of unknowns \( \mathbf{x}_1 \) in the system:
\[
\begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{b}_1 \\ \mathbf{b}_2 \end{bmatrix}
\]
The solution for \( \mathbf{x}_1 \) is:
\[
\mathbf{x}_1 = A^{-1}(\mathbf{b}_1 - B\mathbf{x}_2);
\]
Substituting this into the second block of the equation yields:
\[
\mathbf{b}_2 = C\mathbf{x}_1 + D\mathbf{x}_2 = CA^{-1}(\mathbf{b}_1 - B\mathbf{x}_2) + D\mathbf{x}_2 \implies (D - CA^{-1}B)\mathbf{x}_2 = \mathbf{b}_2 - CA^{-1}\mathbf{b}_1.
\]
**b) Construct the full LU decomposition of \( M \):**
Compute an LU decomposition of \( A \), then evaluate and factorize \( D - CA^{-1}B \). Finally, manipulate the result using efficient operations **without explicitly constructing \( A^{-1} \)**.
This approach ensures computational efficiency while handling large matrix blocks using the principles of linear algebra decomposition techniques.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F153bfb9c-b05e-4b91-94e3-9e151aaf7f28%2Fd43ec349-3056-4847-ac5f-0fc883523c31%2F2i5j4zr_processed.png&w=3840&q=75)
Transcribed Image Text:Given the matrix \( M = \begin{bmatrix} A & B \\ C & D \end{bmatrix} \), where \( A \in \mathbb{R}^{k \times k} \) is nonsingular, we proceed with the following problems:
**a) Verify the "block LU decomposition" formula:**
The block LU decomposition is given by:
\[
M = \begin{bmatrix} I & 0 \\ CA^{-1} & I \end{bmatrix} \begin{bmatrix} A & B \\ 0 & D - CA^{-1}B \end{bmatrix}.
\]
The matrix \( D - CA^{-1}B \) is referred to as the **Schur complement** of \( A \) and is obtained after eliminating the first block of unknowns \( \mathbf{x}_1 \) in the system:
\[
\begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} \mathbf{x}_1 \\ \mathbf{x}_2 \end{bmatrix} = \begin{bmatrix} \mathbf{b}_1 \\ \mathbf{b}_2 \end{bmatrix}
\]
The solution for \( \mathbf{x}_1 \) is:
\[
\mathbf{x}_1 = A^{-1}(\mathbf{b}_1 - B\mathbf{x}_2);
\]
Substituting this into the second block of the equation yields:
\[
\mathbf{b}_2 = C\mathbf{x}_1 + D\mathbf{x}_2 = CA^{-1}(\mathbf{b}_1 - B\mathbf{x}_2) + D\mathbf{x}_2 \implies (D - CA^{-1}B)\mathbf{x}_2 = \mathbf{b}_2 - CA^{-1}\mathbf{b}_1.
\]
**b) Construct the full LU decomposition of \( M \):**
Compute an LU decomposition of \( A \), then evaluate and factorize \( D - CA^{-1}B \). Finally, manipulate the result using efficient operations **without explicitly constructing \( A^{-1} \)**.
This approach ensures computational efficiency while handling large matrix blocks using the principles of linear algebra decomposition techniques.
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