Use the Gram-Schmidt process to find a orthogonal basis for Row (A) where 3 0 3 00 0-2 -2 0 0 -4 -4 0 1 -4 -1 0 1 0 1 0 0 A = 0 0 3 1
Use the Gram-Schmidt process to find a orthogonal basis for Row (A) where 3 0 3 00 0-2 -2 0 0 -4 -4 0 1 -4 -1 0 1 0 1 0 0 A = 0 0 3 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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![**Using the Gram-Schmidt Process to Find an Orthogonal Basis for Row(A)**
In this section, we will perform the Gram-Schmidt process to find an orthogonal basis for the row space of the matrix \( A \). The given matrix \( A \) is:
\[
A =
\begin{bmatrix}
3 & 0 & 3 & 0 & 0 \\
0 & -2 & -2 & 0 & 0 \\
0 & -4 & -4 & 0 & 1 \\
3 & -4 & -1 & 0 & 1 \\
1 & 0 & 1 & 0 & 0
\end{bmatrix}
\]
The row space of a matrix is the set of all linear combinations of its row vectors. To find an orthogonal basis for the row space using the Gram-Schmidt process, we proceed as follows:
1. **Step 1: Select the first vector \(\mathbf{v_1}\)**:
- Let \(\mathbf{v_1}\) be the first row of matrix \( A \).
\[
\mathbf{v_1} = \begin{bmatrix} 3 & 0 & 3 & 0 & 0 \end{bmatrix}
\]
2. **Step 2: Orthogonalize the second row vector \(\mathbf{v_2}\)**:
- Compute the projection of the second row vector \(\mathbf{a_2}\) onto \(\mathbf{v_1}\).
\[
\mathbf{a_2} = \begin{bmatrix} 0 & -2 & -2 & 0 & 0 \end{bmatrix}
\]
- Find \(\mathbf{u_2} = \mathbf{a_2} - \operatorname{proj}_{\mathbf{v_1}} \mathbf{a_2}\).
3. **Step 3: Continue orthogonalizing each subsequent row vector**:
- Apply the Gram-Schmidt process recursively to each row vector to ensure orthogonality with all previously processed vectors:
\[
\mathbf{u_3} = \mathbf{a_3} - \operatorname{proj}_{\mathbf{v_1}} \mathbf{a_3} - \operatorname{proj}_{\mathbf{u_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F644afe59-5ebd-4464-95f6-cbe6de6ea324%2F10c72101-2918-459c-9a82-8e33256ac9ad%2Fexz2si_processed.png&w=3840&q=75)
Transcribed Image Text:**Using the Gram-Schmidt Process to Find an Orthogonal Basis for Row(A)**
In this section, we will perform the Gram-Schmidt process to find an orthogonal basis for the row space of the matrix \( A \). The given matrix \( A \) is:
\[
A =
\begin{bmatrix}
3 & 0 & 3 & 0 & 0 \\
0 & -2 & -2 & 0 & 0 \\
0 & -4 & -4 & 0 & 1 \\
3 & -4 & -1 & 0 & 1 \\
1 & 0 & 1 & 0 & 0
\end{bmatrix}
\]
The row space of a matrix is the set of all linear combinations of its row vectors. To find an orthogonal basis for the row space using the Gram-Schmidt process, we proceed as follows:
1. **Step 1: Select the first vector \(\mathbf{v_1}\)**:
- Let \(\mathbf{v_1}\) be the first row of matrix \( A \).
\[
\mathbf{v_1} = \begin{bmatrix} 3 & 0 & 3 & 0 & 0 \end{bmatrix}
\]
2. **Step 2: Orthogonalize the second row vector \(\mathbf{v_2}\)**:
- Compute the projection of the second row vector \(\mathbf{a_2}\) onto \(\mathbf{v_1}\).
\[
\mathbf{a_2} = \begin{bmatrix} 0 & -2 & -2 & 0 & 0 \end{bmatrix}
\]
- Find \(\mathbf{u_2} = \mathbf{a_2} - \operatorname{proj}_{\mathbf{v_1}} \mathbf{a_2}\).
3. **Step 3: Continue orthogonalizing each subsequent row vector**:
- Apply the Gram-Schmidt process recursively to each row vector to ensure orthogonality with all previously processed vectors:
\[
\mathbf{u_3} = \mathbf{a_3} - \operatorname{proj}_{\mathbf{v_1}} \mathbf{a_3} - \operatorname{proj}_{\mathbf{u_
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