The axioms for a vector space V can be used to prove the elementary properties for a vector space. Because of Axiom 2, Axioms 2 and 4 imply, respectively, that 0 +u = u and -u+u = 0 for all u. Complete the proof that -u is unique by showing that if u + w=0, then w=-u. Use the ten axioms of a vector space to justify each step. Axioms In the following axioms, u, v, and w are in vector space V and c and d are scalars. 1. The sum u + v is in V. 2. u+v=v+u 3. (u+v)+w=u+ (v + w) 4. V has a vector 0 such that u + 0 = u. 5. For each u in V, there is a vector - u in V such that u + (-1)u=0. 6. The scalar multiple cu is in V. 7. c(u + v) = cu + CV 8. (c+d)ucu + du 9. c(du) = (cd)u 10. 1u=u Suppose that w satisfies u + w=0. Adding - u to both sides results in the following. (-u) + [u+w] =(-u) +0 [(-u) +u]+w=(-u) +0 by Axiom 2
The axioms for a vector space V can be used to prove the elementary properties for a vector space. Because of Axiom 2, Axioms 2 and 4 imply, respectively, that 0 +u = u and -u+u = 0 for all u. Complete the proof that -u is unique by showing that if u + w=0, then w=-u. Use the ten axioms of a vector space to justify each step. Axioms In the following axioms, u, v, and w are in vector space V and c and d are scalars. 1. The sum u + v is in V. 2. u+v=v+u 3. (u+v)+w=u+ (v + w) 4. V has a vector 0 such that u + 0 = u. 5. For each u in V, there is a vector - u in V such that u + (-1)u=0. 6. The scalar multiple cu is in V. 7. c(u + v) = cu + CV 8. (c+d)ucu + du 9. c(du) = (cd)u 10. 1u=u Suppose that w satisfies u + w=0. Adding - u to both sides results in the following. (-u) + [u+w] =(-u) +0 [(-u) +u]+w=(-u) +0 by Axiom 2
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
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![### The Axioms for a Vector Space
The axioms for a vector space \( V \) can be used to prove the elementary properties for a vector space. Because of Axiom 2, Axioms 2 and 4 imply, respectively, that \( 0 + u = u \) and \( -u + u = 0 \) for all \( u \). Complete the proof that \( -u \) is unique by showing that if \( u + w = 0 \), then \( w = -u \). Use the ten axioms of a vector space to justify each step.
#### Axioms
In the following axioms, \( u \), \( v \), and \( w \) are in vector space \( V \) and \( c \) and \( d \) are scalars.
1. The sum \( u + v \) is in \( V \).
2. \( u + v = v + u \)
3. \( (u + v) + w = u + (v + w) \)
4. \( V \) has a vector \( 0 \) such that \( u + 0 = u \).
5. For each \( u \) in \( V \), there is a vector \( -u \) in \( V \) such that \( u + (-1)u = 0 \).
6. The scalar multiple \( cu \) is in \( V \).
7. \( (c + v)u = cu + cv \)
8. \( (c + d)u = cu + du \)
9. \( c(du) = (cd)u \)
10. \( 1u = u \)
#### Proof That \( -u \) is Unique
Suppose that \( w \) satisfies \( u + w = 0 \). Adding \( -u \) to both sides results in the following:
\[
(-u) + [u + w] = (-u) + 0
\]
Breaking down the steps further:
\[
[(-u) + u] + w = (-u) + 0
\]
By Axiom 2 (commutativity of addition), we can see:
\[
0 + w = (-u) + 0
\]
Using Axiom 4 (existence of additive identity), we verify:
\[
w = -u
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Transcribed Image Text:### The Axioms for a Vector Space
The axioms for a vector space \( V \) can be used to prove the elementary properties for a vector space. Because of Axiom 2, Axioms 2 and 4 imply, respectively, that \( 0 + u = u \) and \( -u + u = 0 \) for all \( u \). Complete the proof that \( -u \) is unique by showing that if \( u + w = 0 \), then \( w = -u \). Use the ten axioms of a vector space to justify each step.
#### Axioms
In the following axioms, \( u \), \( v \), and \( w \) are in vector space \( V \) and \( c \) and \( d \) are scalars.
1. The sum \( u + v \) is in \( V \).
2. \( u + v = v + u \)
3. \( (u + v) + w = u + (v + w) \)
4. \( V \) has a vector \( 0 \) such that \( u + 0 = u \).
5. For each \( u \) in \( V \), there is a vector \( -u \) in \( V \) such that \( u + (-1)u = 0 \).
6. The scalar multiple \( cu \) is in \( V \).
7. \( (c + v)u = cu + cv \)
8. \( (c + d)u = cu + du \)
9. \( c(du) = (cd)u \)
10. \( 1u = u \)
#### Proof That \( -u \) is Unique
Suppose that \( w \) satisfies \( u + w = 0 \). Adding \( -u \) to both sides results in the following:
\[
(-u) + [u + w] = (-u) + 0
\]
Breaking down the steps further:
\[
[(-u) + u] + w = (-u) + 0
\]
By Axiom 2 (commutativity of addition), we can see:
\[
0 + w = (-u) + 0
\]
Using Axiom 4 (existence of additive identity), we verify:
\[
w = -u
\]
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