Let H= Span{V₁, V₂, V3} and B = {V₁, V₂, V3}. Show that B is a basis for V 1 -7 3 -8 9 -4 7 -4 V3 = -8 6 -9 -4 X = -9 6 - 20 8 and x is in H, and find the B-coordinate vector of x for the given vectors.
Let H= Span{V₁, V₂, V3} and B = {V₁, V₂, V3}. Show that B is a basis for V 1 -7 3 -8 9 -4 7 -4 V3 = -8 6 -9 -4 X = -9 6 - 20 8 and x is in H, and find the B-coordinate vector of x for the given vectors.
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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![### Linear Algebra: Basis and Span Example
#### Problem Statement
Let \( H = \text{Span}\{ \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \} \) and \( B = \{ \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \} \). Show that \( B \) is a basis for \( H \) and \( \mathbf{x} \) is in \( H \), and find the \( B \)-coordinate vector of \( \mathbf{x} \) for the given vectors.
Given vectors:
\[
\mathbf{v_1} = \begin{pmatrix} -7 \\ 3 \\ -8 \\ 6 \end{pmatrix}, \quad \mathbf{v_2} = \begin{pmatrix} 9 \\ -4 \\ 7 \\ -4 \end{pmatrix}, \quad \mathbf{v_3} = \begin{pmatrix} -8 \\ 6 \\ -9 \\ -4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} -9 \\ 6 \\ -20 \\ 8 \end{pmatrix}
\]
#### Matrix Representation
Reduce the augmented matrix \([ \mathbf{v_1} \ \mathbf{v_2} \ \mathbf{v_3} \ \mathbf{x} ]\) to reduced echelon form.
\[
\begin{bmatrix}
-7 & 9 & -8 & -9 \\
3 & -4 & 6 & 6 \\
-8 & 7 & -9 & -20 \\
6 & -4 & -4 & 8
\end{bmatrix}
\sim
\begin{bmatrix}
\boxed{}
\end{bmatrix}
\]
#### Questions
1. How can it be shown that \( B \) is a basis for \( H \)?
- A. The first three columns of the augmented matrix are pivot columns and therefore \( B \) forms a basis for \( H \).
- B. The augmented matrix is upper triangular and row equivalent to \([B \ | \mathbf{x} ]\) therefore \( B \) forms a basis for \( H \).
- C. The augmented matrix shows that the system of equations is consistent and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2Fb257a014-71b9-46d0-9883-46afbe8b7ed8%2F84f87ep_processed.png&w=3840&q=75)
Transcribed Image Text:### Linear Algebra: Basis and Span Example
#### Problem Statement
Let \( H = \text{Span}\{ \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \} \) and \( B = \{ \mathbf{v_1}, \mathbf{v_2}, \mathbf{v_3} \} \). Show that \( B \) is a basis for \( H \) and \( \mathbf{x} \) is in \( H \), and find the \( B \)-coordinate vector of \( \mathbf{x} \) for the given vectors.
Given vectors:
\[
\mathbf{v_1} = \begin{pmatrix} -7 \\ 3 \\ -8 \\ 6 \end{pmatrix}, \quad \mathbf{v_2} = \begin{pmatrix} 9 \\ -4 \\ 7 \\ -4 \end{pmatrix}, \quad \mathbf{v_3} = \begin{pmatrix} -8 \\ 6 \\ -9 \\ -4 \end{pmatrix}, \quad \mathbf{x} = \begin{pmatrix} -9 \\ 6 \\ -20 \\ 8 \end{pmatrix}
\]
#### Matrix Representation
Reduce the augmented matrix \([ \mathbf{v_1} \ \mathbf{v_2} \ \mathbf{v_3} \ \mathbf{x} ]\) to reduced echelon form.
\[
\begin{bmatrix}
-7 & 9 & -8 & -9 \\
3 & -4 & 6 & 6 \\
-8 & 7 & -9 & -20 \\
6 & -4 & -4 & 8
\end{bmatrix}
\sim
\begin{bmatrix}
\boxed{}
\end{bmatrix}
\]
#### Questions
1. How can it be shown that \( B \) is a basis for \( H \)?
- A. The first three columns of the augmented matrix are pivot columns and therefore \( B \) forms a basis for \( H \).
- B. The augmented matrix is upper triangular and row equivalent to \([B \ | \mathbf{x} ]\) therefore \( B \) forms a basis for \( H \).
- C. The augmented matrix shows that the system of equations is consistent and
![**Vector Spaces and Bases**
---
**Problem Statement:**
Let \( H = \text{Span}\{ v_1, v_2, v_3 \} \) and \( B = \{ v_1, v_2, v_3 \} \). Show that \( B \) is a basis for \( H \) and \( x \) is in \( H \), and find the \( B \)-coordinate vector of \( x \) for the given vectors.
\[
v_1 = \begin{bmatrix} -7 \\ 3 \\ -8 \\ 6 \end{bmatrix} ,
v_2 = \begin{bmatrix} 9 \\ -4 \\ 7 \\ -4 \end{bmatrix} ,
v_3 = \begin{bmatrix} -8 \\ 6 \\ -9 \\ -4 \end{bmatrix} ,
x = \begin{bmatrix} -9 \\ 6 \\ -20 \\ 8 \end{bmatrix}
\]
---
**Discussion:**
1. **How can it be shown that \( B \) is a basis for \( H \)?**
- **A.** The first three columns of the augmented matrix are pivot columns and therefore \( B \) forms a basis for \( H \).
- **B.** The augmented matrix is upper triangular and row equivalent to
\[
\left[ B \quad x \right]
\]
therefore, \( B \) forms a basis for \( H \).
- **C.** The augmented matrix shows that the system of equations is consistent and therefore \( B \) forms a basis for \( H \).
- **D.** \( H \) is the Span{\( v_1, v_2, v_3 \)} and \( B = \{ v_1, v_2, v_3 \} \) so therefore \( B \) must form a basis for \( H \).
2. **How can it be shown that \( x \) is in \( H \)?**
- **A.** The augmented matrix is upper triangular and row equivalent to
\[
\left[ B \quad x \right]
\]
therefore \( x \) is in \( H \) because \( H \) is the Span{\( v_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F70fe4f99-9e69-4bef-afe6-80d54e8903ee%2Fb257a014-71b9-46d0-9883-46afbe8b7ed8%2F24l22l_processed.png&w=3840&q=75)
Transcribed Image Text:**Vector Spaces and Bases**
---
**Problem Statement:**
Let \( H = \text{Span}\{ v_1, v_2, v_3 \} \) and \( B = \{ v_1, v_2, v_3 \} \). Show that \( B \) is a basis for \( H \) and \( x \) is in \( H \), and find the \( B \)-coordinate vector of \( x \) for the given vectors.
\[
v_1 = \begin{bmatrix} -7 \\ 3 \\ -8 \\ 6 \end{bmatrix} ,
v_2 = \begin{bmatrix} 9 \\ -4 \\ 7 \\ -4 \end{bmatrix} ,
v_3 = \begin{bmatrix} -8 \\ 6 \\ -9 \\ -4 \end{bmatrix} ,
x = \begin{bmatrix} -9 \\ 6 \\ -20 \\ 8 \end{bmatrix}
\]
---
**Discussion:**
1. **How can it be shown that \( B \) is a basis for \( H \)?**
- **A.** The first three columns of the augmented matrix are pivot columns and therefore \( B \) forms a basis for \( H \).
- **B.** The augmented matrix is upper triangular and row equivalent to
\[
\left[ B \quad x \right]
\]
therefore, \( B \) forms a basis for \( H \).
- **C.** The augmented matrix shows that the system of equations is consistent and therefore \( B \) forms a basis for \( H \).
- **D.** \( H \) is the Span{\( v_1, v_2, v_3 \)} and \( B = \{ v_1, v_2, v_3 \} \) so therefore \( B \) must form a basis for \( H \).
2. **How can it be shown that \( x \) is in \( H \)?**
- **A.** The augmented matrix is upper triangular and row equivalent to
\[
\left[ B \quad x \right]
\]
therefore \( x \) is in \( H \) because \( H \) is the Span{\( v_
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