(a) (b) Any set of vectors containing the zero vector is linearly dependent. Any set of more than ʼn vectors in R" is linearly dependent.
(a) (b) Any set of vectors containing the zero vector is linearly dependent. Any set of more than ʼn vectors in R" is linearly dependent.
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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![### Definition and Exploration of Linear Combinations, Linear Dependence, and Linear Independence
#### Key Concepts:
- **Linear Combination**: A linear combination of vectors a, b, and c involves an expression of the form αa + βb + γc, where α, β, and γ are scalars.
- **Linear Dependence**: Vectors a, b, and c are linearly dependent if there exist scalars, not all zero, such that their linear combination equals the zero vector.
- **Linear Independence**: Vectors are linearly independent if the only scalars that satisfy the linear combination resulting in the zero vector are all zero.
#### Problem Overview:
Consider three vectors \( a, b, \) and \( c \) in \( \mathbb{R}^n \).
1. **Statement (i):** If \( a, b, \) and \( c \) are linearly dependent, then there exist scalars \( \alpha, \beta, \) and \( \gamma \), not all zero, such that:
\[
\alpha a + \beta b + \gamma c = 0
\]
2. **Statement (ii):** Conversely, if there exist scalars \( \alpha, \beta, \) and \( \gamma \), not all zero, such that:
\[
\alpha a + \beta b + \gamma c = 0
\]
then \( a, b, \) and \( c \) are linearly dependent.
#### Generalization:
Extend this principle to \( k \) vectors in \( \mathbb{R}^n \). This leads to the conclusion that for any set of vectors in such a space, linear dependence can similarly be demonstrated using scalar multiplication.
#### Application:
Using the above generalization, prove the following statements:
(a) **Zero Vector Dependence:**
Any set of vectors that includes the zero vector is linearly dependent because any scalar multiplied by the zero vector is still zero, leading to a nontrivial linear combination to equal zero.
(b) **Dependence Beyond Dimensions:**
Any set containing more than \( n \) vectors in \( \mathbb{R}^n \) is linearly dependent. This is because the vectors exceed the dimension they are within, making it impossible for all combinations to remain independent. This aligns with the classic theorem regarding dimensions and vector spaces](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffba66248-862a-43fb-91b6-16cc301cecb0%2F3f14cb10-6037-446a-84dd-e9cb691bf463%2Fiuf37b_processed.png&w=3840&q=75)
Transcribed Image Text:### Definition and Exploration of Linear Combinations, Linear Dependence, and Linear Independence
#### Key Concepts:
- **Linear Combination**: A linear combination of vectors a, b, and c involves an expression of the form αa + βb + γc, where α, β, and γ are scalars.
- **Linear Dependence**: Vectors a, b, and c are linearly dependent if there exist scalars, not all zero, such that their linear combination equals the zero vector.
- **Linear Independence**: Vectors are linearly independent if the only scalars that satisfy the linear combination resulting in the zero vector are all zero.
#### Problem Overview:
Consider three vectors \( a, b, \) and \( c \) in \( \mathbb{R}^n \).
1. **Statement (i):** If \( a, b, \) and \( c \) are linearly dependent, then there exist scalars \( \alpha, \beta, \) and \( \gamma \), not all zero, such that:
\[
\alpha a + \beta b + \gamma c = 0
\]
2. **Statement (ii):** Conversely, if there exist scalars \( \alpha, \beta, \) and \( \gamma \), not all zero, such that:
\[
\alpha a + \beta b + \gamma c = 0
\]
then \( a, b, \) and \( c \) are linearly dependent.
#### Generalization:
Extend this principle to \( k \) vectors in \( \mathbb{R}^n \). This leads to the conclusion that for any set of vectors in such a space, linear dependence can similarly be demonstrated using scalar multiplication.
#### Application:
Using the above generalization, prove the following statements:
(a) **Zero Vector Dependence:**
Any set of vectors that includes the zero vector is linearly dependent because any scalar multiplied by the zero vector is still zero, leading to a nontrivial linear combination to equal zero.
(b) **Dependence Beyond Dimensions:**
Any set containing more than \( n \) vectors in \( \mathbb{R}^n \) is linearly dependent. This is because the vectors exceed the dimension they are within, making it impossible for all combinations to remain independent. This aligns with the classic theorem regarding dimensions and vector spaces
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