Use Axioms only to prove that the the solution to the equation 2x + y + 3z = 0 is a subspace of R³.
Use Axioms only to prove that the the solution to the equation 2x + y + 3z = 0 is a subspace of R³.
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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### Proving Subspaces Using Axioms
**Problem Statement:**
Use Axioms only to prove that the solution to the equation \(2x + y + 3z = 0\) is a subspace of \(\mathbb{R}^3\).
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In this problem, we need to show that the set of all solutions to the given linear equation forms a subspace of \(\mathbb{R}^3\). The set of solutions can be denoted as:
\[ S = \{ (x, y, z) \in \mathbb{R}^3 \mid 2x + y + 3z = 0 \} \]
To prove that \( S \) is a subspace, we need to verify that it satisfies the three main subspace axioms:
1. **Zero Vector Presence**: The zero vector of \(\mathbb{R}^3\) (which is (0,0,0)) must be in \(S\).
2. **Closure Under Addition**: For any two vectors \(u\) and \(v\) in \(S\), the vector \(u + v\) must also be in \(S\).
3. **Closure Under Scalar Multiplication**: For any vector \(u\) in \(S\) and any scalar \(c\), the vector \(cu\) must also be in \(S\).
---
### Verification
#### 1. Zero Vector
The zero vector in \(\mathbb{R}^3\) is \((0, 0, 0)\). We check if it satisfies the given equation:
\[ 2(0) + (0) + 3(0) = 0 \]
Since this holds true, the zero vector is in \( S \).
#### 2. Closure Under Addition
Let \( u = (x_1, y_1, z_1) \in S \) and \( v = (x_2, y_2, z_2) \in S \). Then by definition:
\[ 2x_1 + y_1 + 3z_1 = 0 \quad \text{and} \quad 2x_2 + y_2 + 3z_2 = 0 \]
We need to show that \( u + v = (x_1 + x_](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5483afab-7850-40f6-959f-e698dea1419a%2F14bde62f-8dcd-4541-88e8-bb59c05ee0a7%2F8ef6gpp_processed.png&w=3840&q=75)
Transcribed Image Text:---
### Proving Subspaces Using Axioms
**Problem Statement:**
Use Axioms only to prove that the solution to the equation \(2x + y + 3z = 0\) is a subspace of \(\mathbb{R}^3\).
---
In this problem, we need to show that the set of all solutions to the given linear equation forms a subspace of \(\mathbb{R}^3\). The set of solutions can be denoted as:
\[ S = \{ (x, y, z) \in \mathbb{R}^3 \mid 2x + y + 3z = 0 \} \]
To prove that \( S \) is a subspace, we need to verify that it satisfies the three main subspace axioms:
1. **Zero Vector Presence**: The zero vector of \(\mathbb{R}^3\) (which is (0,0,0)) must be in \(S\).
2. **Closure Under Addition**: For any two vectors \(u\) and \(v\) in \(S\), the vector \(u + v\) must also be in \(S\).
3. **Closure Under Scalar Multiplication**: For any vector \(u\) in \(S\) and any scalar \(c\), the vector \(cu\) must also be in \(S\).
---
### Verification
#### 1. Zero Vector
The zero vector in \(\mathbb{R}^3\) is \((0, 0, 0)\). We check if it satisfies the given equation:
\[ 2(0) + (0) + 3(0) = 0 \]
Since this holds true, the zero vector is in \( S \).
#### 2. Closure Under Addition
Let \( u = (x_1, y_1, z_1) \in S \) and \( v = (x_2, y_2, z_2) \in S \). Then by definition:
\[ 2x_1 + y_1 + 3z_1 = 0 \quad \text{and} \quad 2x_2 + y_2 + 3z_2 = 0 \]
We need to show that \( u + v = (x_1 + x_
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