A singular value decomposition of A is as follows: 0.5 0.5 0.5 -0.5 -0.5 0.5 0.5 -0.5 -0.5 -0.5 0.5 0.5] [5 01 0.5 0.5 -0.5 0.5 Find the least-squares solution of the linear system -1 A=UEVT Ax = b, where b â₁ Ĵ₂ = - 3 2 4 0 5 [0.6 0 0 0.8 0 0 -0.8] 0.6

Advanced Engineering Mathematics
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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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**Least-Squares Solution Using Singular Value Decomposition (SVD)**

**Singular Value Decomposition:**

A singular value decomposition (SVD) of \( A \) is given as follows:

\[ 
A = U \Sigma V^T = 
\begin{bmatrix}
    0.5 & 0.5 & 0.5 & 0.5 \\
    -0.5 & -0.5 & 0.5 & 0.5 \\
    0.5 & -0.5 & 0.5 & -0.5 \\
    -0.5 & 0.5 & -0.5 & 0.5 
\end{bmatrix}
\begin{bmatrix}
    5 & 0 \\
    0 & 5 \\
    0 & 0 \\
    0 & 0 
\end{bmatrix}
\begin{bmatrix}
    0.6 & -0.8 \\
    0.8 & 0.6
\end{bmatrix}
\]

**Objective:**

Find the least-squares solution of the linear system:

\[ Ax = b, \text{ where } b = 
\begin{bmatrix}
    -1 \\
    3 \\
    2 \\
    4 
\end{bmatrix} 
\]

**Solutions:**

\[ 
\hat{x}_1 = \; \_\_\_;
\]
\[ 
\hat{x}_2 = \; \_\_\_;
\]

By solving the system using the given SVD of the matrix \( A \), we can find the least-squares solution \( \hat{x} \).

**Explanation of Graphs/Diagrams:**

Here we have a matrix \( A \) decomposed into three matrices \( U \), \( \Sigma \), and \( V^T \). 
- \( U \) is a 4x4 orthogonal matrix.
- \( \Sigma \) is a 4x2 diagonal matrix with non-negative real numbers on the diagonal (the singular values).
- \( V^T \) is the transpose of a 2x2 orthogonal matrix.

Using this decomposition, we find \( x \) that minimizes the equation \( Ax = b \).
Transcribed Image Text:**Least-Squares Solution Using Singular Value Decomposition (SVD)** **Singular Value Decomposition:** A singular value decomposition (SVD) of \( A \) is given as follows: \[ A = U \Sigma V^T = \begin{bmatrix} 0.5 & 0.5 & 0.5 & 0.5 \\ -0.5 & -0.5 & 0.5 & 0.5 \\ 0.5 & -0.5 & 0.5 & -0.5 \\ -0.5 & 0.5 & -0.5 & 0.5 \end{bmatrix} \begin{bmatrix} 5 & 0 \\ 0 & 5 \\ 0 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0.6 & -0.8 \\ 0.8 & 0.6 \end{bmatrix} \] **Objective:** Find the least-squares solution of the linear system: \[ Ax = b, \text{ where } b = \begin{bmatrix} -1 \\ 3 \\ 2 \\ 4 \end{bmatrix} \] **Solutions:** \[ \hat{x}_1 = \; \_\_\_; \] \[ \hat{x}_2 = \; \_\_\_; \] By solving the system using the given SVD of the matrix \( A \), we can find the least-squares solution \( \hat{x} \). **Explanation of Graphs/Diagrams:** Here we have a matrix \( A \) decomposed into three matrices \( U \), \( \Sigma \), and \( V^T \). - \( U \) is a 4x4 orthogonal matrix. - \( \Sigma \) is a 4x2 diagonal matrix with non-negative real numbers on the diagonal (the singular values). - \( V^T \) is the transpose of a 2x2 orthogonal matrix. Using this decomposition, we find \( x \) that minimizes the equation \( Ax = b \).
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