Find the orthogonal complement S¹, and find the direct sum S + S¹. 6 -08 S = span 1 1 (a) Find the orthogonal complement St. S² = span
Find the orthogonal complement S¹, and find the direct sum S + S¹. 6 -08 S = span 1 1 (a) Find the orthogonal complement St. S² = span
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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![### Finding the Orthogonal Complement and Direct Sum in Linear Algebra
This tutorial will guide you through finding the orthogonal complement \( S^{\perp} \) of a subspace \( S \), and computing the direct sum \( S \oplus S^{\perp} \).
Given:
\[ S = \text{span}\left\{\begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
6 \\ 0 \\ 1
\end{bmatrix}\right\} \]
#### (a) Finding the Orthogonal Complement \( S^{\perp} \)
To find \( S^{\perp} \), we need to identify all vectors orthogonal to every vector in \( S \).
We use the concept that \( S^{\perp} \) is the set of all vectors that are orthogonal to each vector in \( S \). We form the equation \( A \mathbf{x} = 0 \), where \( A \) is the matrix whose rows are the given vectors in \( S \):
\[ A = \begin{bmatrix}
0 & 1 & 0 \\
6 & 0 & 1
\end{bmatrix} \]
To find the null space of \( A \), solve \( A \mathbf{x} = 0 \).
This is depicted in the image with three blank boxes representing the components of a general vector in \( \mathbb{R}^3 \). Arrows indicate that we need to form combined linear equations from the vectors of \( S \), solve, and then extract the spanning vectors for \( S^{\perp} \).
\[ S^{\perp} = \text{span}\left\{
\begin{bmatrix}
\text{[Solution1]} \\
\text{[Solution2]} \\
\text{[Solution3]}
\end{bmatrix}
\right\} \]
Finally, take the direct sum \( S \oplus S^{\perp} \) which should span the entire space \( \mathbb{R}^3 \).
Use this approach to confirm if the obtained vectors indeed form an orthogonal basis.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F031412a4-ce24-4200-b630-467aaa8d4c51%2F7c753fe2-f6af-4024-a747-257e035fc0e4%2Ftvlfnyl_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Finding the Orthogonal Complement and Direct Sum in Linear Algebra
This tutorial will guide you through finding the orthogonal complement \( S^{\perp} \) of a subspace \( S \), and computing the direct sum \( S \oplus S^{\perp} \).
Given:
\[ S = \text{span}\left\{\begin{bmatrix}
0 \\ 1 \\ 0
\end{bmatrix},
\begin{bmatrix}
6 \\ 0 \\ 1
\end{bmatrix}\right\} \]
#### (a) Finding the Orthogonal Complement \( S^{\perp} \)
To find \( S^{\perp} \), we need to identify all vectors orthogonal to every vector in \( S \).
We use the concept that \( S^{\perp} \) is the set of all vectors that are orthogonal to each vector in \( S \). We form the equation \( A \mathbf{x} = 0 \), where \( A \) is the matrix whose rows are the given vectors in \( S \):
\[ A = \begin{bmatrix}
0 & 1 & 0 \\
6 & 0 & 1
\end{bmatrix} \]
To find the null space of \( A \), solve \( A \mathbf{x} = 0 \).
This is depicted in the image with three blank boxes representing the components of a general vector in \( \mathbb{R}^3 \). Arrows indicate that we need to form combined linear equations from the vectors of \( S \), solve, and then extract the spanning vectors for \( S^{\perp} \).
\[ S^{\perp} = \text{span}\left\{
\begin{bmatrix}
\text{[Solution1]} \\
\text{[Solution2]} \\
\text{[Solution3]}
\end{bmatrix}
\right\} \]
Finally, take the direct sum \( S \oplus S^{\perp} \) which should span the entire space \( \mathbb{R}^3 \).
Use this approach to confirm if the obtained vectors indeed form an orthogonal basis.
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