20. Prove that a vector u in a vector space has only ative, -u. one neg-
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![### Problem 20
**Prove that a vector** **u** **in a vector space has only one negative, \(-\mathbf{u}\).**
#### Explanation
In any vector space, the negative of a vector \( \mathbf{u} \) is defined such that when it is added to the original vector, the result is the zero vector, \( \mathbf{0} \). Mathematically, this is expressed as:
\[ \mathbf{u} + (-\mathbf{u}) = \mathbf{0} \]
To prove that a vector has only one negative, assume there exists another vector \( \mathbf{v} \) such that:
\[ \mathbf{u} + \mathbf{v} = \mathbf{0} \]
If both \( -\mathbf{u} \) and \( \mathbf{v} \) satisfy this property, substitute \( \mathbf{v} = -\mathbf{u} \), showing that:
\[ -\mathbf{u} + \mathbf{v} = \mathbf{0} \]
Thus, the only vector that can negate \( \mathbf{u} \) to result in the zero vector is \( -\mathbf{u} \). Consequently, this proves the uniqueness of the negative vector for any given vector in a vector space.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F832196dd-2d45-47e5-91ed-72a7273417b1%2F5377caf5-202e-4b6c-b00c-398181bfe91a%2Fgwtfjsa_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 20
**Prove that a vector** **u** **in a vector space has only one negative, \(-\mathbf{u}\).**
#### Explanation
In any vector space, the negative of a vector \( \mathbf{u} \) is defined such that when it is added to the original vector, the result is the zero vector, \( \mathbf{0} \). Mathematically, this is expressed as:
\[ \mathbf{u} + (-\mathbf{u}) = \mathbf{0} \]
To prove that a vector has only one negative, assume there exists another vector \( \mathbf{v} \) such that:
\[ \mathbf{u} + \mathbf{v} = \mathbf{0} \]
If both \( -\mathbf{u} \) and \( \mathbf{v} \) satisfy this property, substitute \( \mathbf{v} = -\mathbf{u} \), showing that:
\[ -\mathbf{u} + \mathbf{v} = \mathbf{0} \]
Thus, the only vector that can negate \( \mathbf{u} \) to result in the zero vector is \( -\mathbf{u} \). Consequently, this proves the uniqueness of the negative vector for any given vector in a vector space.
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
