Solutions to Linear Algebra Done Right Third Edition
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Chapter 1 Exercise B Posted on January 2, 2016 by Mohammad Rashidi
1. Prove that −(−v) = v for every v ∈ V . Solution: By definition, we have
(−v) + (−(−v)) = 0 an and d
v +
(−v) = 0. (−
This implies both v and −(−v) are additive inverses of −v , by the uniqueness of additive inverse, it follows that −(−v) = v .
2. Suppose a ∈ F , v ∈ V , and av = 0 . Prove that a = 0 or v = 0 . Solution: If a ≠ 0 , then a has inverse a−1 such that a−1 a = 1 . Hence v =
1 ⋅ v = (a−1 a)v = a−1 (av) = a−1 ⋅ 0 = 0.
Here we use associativity in 1.19 and and 1.30.
3. Suppose v, w ∈ V . Explain why there exists a unique x ∈ V such that v + 3x = w . Solution: Let x =
1 (w − v ) , then 3 v + 3x = v
+3⋅
1 (w − v ) = v + (w − v) = w . 3
This shows shows existence. Now we show uniqueness. Suppose, we have have another another vector x′ such that ′ v + 3x
= w . Then v + 3x′ = w implies 3x′ = w − v . Similarly, 3x = w − v . Hence 3(x − x′ ) = 3x − 3x′ = (w − v ) − (w − v ) = 0.
By Problem 2, it follows that x − x′ = 0 . This shows uniqueness.
4. The empty set is not a vector space. The empty set fails to satisfy only one of the requirements listed in 1.19. Which one? Solution: Additive identity: there exists an element 0 ∈ V such that v + 0 = v for all v ∈ V ; This means V cannot be empty.
5. Show that in the definition of a vector space (1.19), the additive inverse condition can be replaced with the condition that
0v = 0 for all v ∈ V .
(1)
Here the 0 on the left side is the number 0 , and the 0 on the right side is the additive identity of V . (The phrase “a condition can be replaced” in a definition means that the collection of objects satisfying the definition is unchanged if the original condition is replaced with the new condition.) Solution: If we assume the additive inverse condition, we already show (1) in 1.29. Now we assume (1) and then show additive inverse condition. Since we have (1) , we have v +
((−1)v) = 1v + ((−1)v) = (1 + (−1))v = 0v = 0,
this means the existence of additive inverse, i.e. the additive inverse condition.
6. Let ∞ and −∞ denote two distinct objects, neither of which is in R . Define an addition and scalar multiplication on R ∪ {∞} ∪ {∞} as you could guess from the notation. Specifically, the sum and product of two real numbers is as usual, and for t ∈ R define
⎪⎧ −∞, ∞ = ⎨ 0, ⎪⎩ ∞,
t
t +
if t < 0, if t = 0, if t > 0,
∞ = ∞ + t = ∞,
∞ + ∞ = ∞,
⎪⎧ ∞, (−∞) = ⎨ 0, ⎪⎩ −∞,
t
if t < 0, if t = 0, if t > 0,
t +
(−∞) = (−∞) + t = −∞,
(−∞) + (−∞) = −∞,
∞ + (−∞) = 0.
Is R ∪ {∞} ∪ {∞} a vector space over R ? Explain. Solution: This is not a vector space over R . Consider the distributive properties in 1.19. If this is a vector space over R , we will have
∞ = (2 + (−1))∞ = 2∞ + (−1)∞ = ∞ + (−∞) = 0.
Hence for any t ∈ R , one has t =
0 + t = ∞ + t = ∞ = 0.
We get a contradiction since zero vector is unique.
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6 Comments Recommend
Solutions to Linear Algebra Done Right 3rd Edition 11
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Join the discussion… MASHIAT MUTMAINNAH • a year ago
Hi! Thank you so much for all the solutions, I really appreciate your hard work. Can someone please explain to me the contradiction in the solution Ex 1B question 6? I just can't seem to understand it. Thank you! 1
• Reply • Share › Mohammad Rashidi
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> MASHIAT MUTMAINNAH • 5 months ago
Since zero vector should be unique. • Reply • Share › Abu-Yakitori > Mohammad Rashidi • 3 months ago
Why 0 + t = infinity + t? • Reply • Share › spindash > Abu-Yakitori • 3 months ago
From the first equation we obtain ∞ = 0 (only using distributive properties). Whereas in the second equation we get t = 0 That is the contradiction, 0 is not unique. 0+t = ∞+t since by the result of the first equation ∞ = 0. So if 0=∞, 0+t=∞+t. 1
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Maxis Jaisi • 2 months ago
Alternatively, we could just observe that $ t + \infty = \infty$ for every $ t \in \mathbf{R}$ automatically violates uniqueness of additive identity. • Reply • Share › Rong Ou • 2 months ago
For 6 it also violates associativity: (\infty + \infty) + (-\infty) = \infty + (-\infty) = 0 \infty + (\infty + (-\infty)) = \infty + 0 = \infty • Reply • Share ›
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