What is wrong with the following "proof" that a0 = = 0, where 0 is the zero in a vector space V? a0 = 0 for any scalar a € F. Proof Since zero multiplied by anything equals zero, we have that a0 = 0.
What is wrong with the following "proof" that a0 = = 0, where 0 is the zero in a vector space V? a0 = 0 for any scalar a € F. Proof Since zero multiplied by anything equals zero, we have that a0 = 0.
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
![**Question:**
What is wrong with the following “proof” that \( \alpha \mathbf{0} = \mathbf{0} \), where \( \mathbf{0} \) is the zero in a vector space \( V \)?
\[
\alpha \mathbf{0} = \mathbf{0} \text{ for any scalar } \alpha \in \mathbb{F}.
\]
**Proof:**
Since zero multiplied by anything equals zero, we have that \( \alpha \mathbf{0} = \mathbf{0} \).
∎
**Discussion:**
The question challenges the validity of the proof that attempts to show that multiplying the zero vector \( \mathbf{0} \) by any scalar \( \alpha \) in a vector space always results in the zero vector, based on a property of multiplication in arithmetic not applicable in this context without further axiomatic justification specific to vector spaces.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb04829d0-4645-426e-bf1a-7ada40b0786f%2F0aa85ca9-de05-434f-8baa-481af0d0cd40%2Fuc3og7w_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
What is wrong with the following “proof” that \( \alpha \mathbf{0} = \mathbf{0} \), where \( \mathbf{0} \) is the zero in a vector space \( V \)?
\[
\alpha \mathbf{0} = \mathbf{0} \text{ for any scalar } \alpha \in \mathbb{F}.
\]
**Proof:**
Since zero multiplied by anything equals zero, we have that \( \alpha \mathbf{0} = \mathbf{0} \).
∎
**Discussion:**
The question challenges the validity of the proof that attempts to show that multiplying the zero vector \( \mathbf{0} \) by any scalar \( \alpha \) in a vector space always results in the zero vector, based on a property of multiplication in arithmetic not applicable in this context without further axiomatic justification specific to vector spaces.
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 3 steps with 2 images
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,
