Use the least squares method to find the orthogonal projection of b = [1 -2 2] onto the column space of the matrix A. A = 12 projs b = 2/3 -5/3 -7/3 ↓ ↑
Use the least squares method to find the orthogonal projection of b = [1 -2 2] onto the column space of the matrix A. A = 12 projs b = 2/3 -5/3 -7/3 ↓ ↑
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
![**Orthogonal Projection Using the Least Squares Method**
**Problem Statement:**
Use the least squares method to find the orthogonal projection of the vector \( \mathbf{b} = \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \) onto the column space of the matrix \( A \).
The matrix \( A \) is given as:
\[
A = \begin{bmatrix}
1 & 1 \\
1 & 2 \\
0 & 1
\end{bmatrix}
\]
**Solution:**
To find the orthogonal projection of \( \mathbf{b} \) onto the column space of \( A \), we use the formula for the projection:
\[
\text{proj}_{S} \, \mathbf{b} = \begin{bmatrix} \frac{2}{3} \\ -\frac{5}{3} \\ -\frac{7}{3} \end{bmatrix}
\]
**Explanation of Diagram:**
The diagram includes the vector \( \text{proj}_{S} \, \mathbf{b} \), represented in a column format:
\[
\text{proj}_{S} \, \mathbf{b} =
\begin{bmatrix}
2/3 \\
-5/3 \\
-7/3
\end{bmatrix}
\]
Arrows on the sides and bottom of the components indicate directions for vector manipulation or transformation, visually suggesting movement or transition within the space defined by matrix \( A \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8b8a4ad9-b66e-4dbe-8b3f-39bf0adb3cae%2F43a212b3-6716-448b-8e0c-2535841491f9%2F8p8m2h_processed.png&w=3840&q=75)
Transcribed Image Text:**Orthogonal Projection Using the Least Squares Method**
**Problem Statement:**
Use the least squares method to find the orthogonal projection of the vector \( \mathbf{b} = \begin{bmatrix} 1 \\ -2 \\ 2 \end{bmatrix} \) onto the column space of the matrix \( A \).
The matrix \( A \) is given as:
\[
A = \begin{bmatrix}
1 & 1 \\
1 & 2 \\
0 & 1
\end{bmatrix}
\]
**Solution:**
To find the orthogonal projection of \( \mathbf{b} \) onto the column space of \( A \), we use the formula for the projection:
\[
\text{proj}_{S} \, \mathbf{b} = \begin{bmatrix} \frac{2}{3} \\ -\frac{5}{3} \\ -\frac{7}{3} \end{bmatrix}
\]
**Explanation of Diagram:**
The diagram includes the vector \( \text{proj}_{S} \, \mathbf{b} \), represented in a column format:
\[
\text{proj}_{S} \, \mathbf{b} =
\begin{bmatrix}
2/3 \\
-5/3 \\
-7/3
\end{bmatrix}
\]
Arrows on the sides and bottom of the components indicate directions for vector manipulation or transformation, visually suggesting movement or transition within the space defined by matrix \( A \).
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