Below is a seven-step proof of the following theorem. Specify which of the ten vector space axioms applies. Theorem: Let u be any vector in a vector space V, let O be the zero vector in V, and let k be a scalar. Then k0=0. Proof: Reasons: (1) k0+ ku k(0+u) (2) - ku (3) Since ku is in V, -ku is in V. (4) Therefore, (k0+ku) + (-ku) ku + (-ku). Add ku to both sides (5) k0+ (ku+(-ku))-ku + (-ku) (6) k0+0=0 (7) k0=0

Elementary Linear Algebra (MindTap Course List)
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Chapter4: Vector Spaces
Section4.4: Spanning Sets And Linear Independence
Problem 74E: Let u, v, and w be any three vectors from a vector space V. Determine whether the set of vectors...
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Below is a seven-step proof of the following theorem. Specify which of the ten vector space axioms applies.
Theorem: Let u be any vector in a vector space V, let 0 be the zero vector in V, and let k be a scalar.
Then k0 = 0.
Proof:
Reasons:
(1) k0+ kuk(0+u)
(2)
= ku
(3) Since ku is in V, -ku is in V.
(4) Therefore, (k0+ku) + (ku)=ku + (-ku).
Add ku to both sides
(5) k0 + (ku + (ku)) = ku + (-ku)
(6) k0+0=0
(7) k0=0
Transcribed Image Text:Below is a seven-step proof of the following theorem. Specify which of the ten vector space axioms applies. Theorem: Let u be any vector in a vector space V, let 0 be the zero vector in V, and let k be a scalar. Then k0 = 0. Proof: Reasons: (1) k0+ kuk(0+u) (2) = ku (3) Since ku is in V, -ku is in V. (4) Therefore, (k0+ku) + (ku)=ku + (-ku). Add ku to both sides (5) k0 + (ku + (ku)) = ku + (-ku) (6) k0+0=0 (7) k0=0
A. closure under addition
B. vector addition is commutative
C. vector addition is associative
D. additive identity
E. additive inverse
F. closure under scalar multiplication
G. a (uv) = a Oua Ov
for any u, v, € V and any real number a
H. (a + b) Ou=aOub Qu
for any u € V and any real numbers a and b
I. (ab) Ou= a (bu) for any u € V and
any real numbers a and b
J. 1 Ou = u for any u E V
Transcribed Image Text:A. closure under addition B. vector addition is commutative C. vector addition is associative D. additive identity E. additive inverse F. closure under scalar multiplication G. a (uv) = a Oua Ov for any u, v, € V and any real number a H. (a + b) Ou=aOub Qu for any u € V and any real numbers a and b I. (ab) Ou= a (bu) for any u € V and any real numbers a and b J. 1 Ou = u for any u E V
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