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Problem

Let $V = \{(a_1, a_2): a_1, a_2 \in F\}$ where $F$ is a field. Define addition of elements of $V$ coordinatewise, and for $c \in F$ and $(a_1,a_2) \in V$, define

$c(a_1, a_2) = (a_1, 0)$

Is $V$ a vector space over $F$ with these operations?

My Solution

$0 \not \in V$; therefore, V is not a vector space since it is not closed under scalar multiplication.

Textbook Solution

No. Since $0(a_1, a_2) = (a_1, 0)$ is the zero vector but this will make the zero vector not be unique, it cannot be a vector space.

I do not understand the textbook's solution and would greatly appreciate it if people could please take the time to review mine. If there is anything wrong with my reasoning, please specify why and what the correct solution is.

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    Your definition of scalar multiplication isn't correct: where is "c" ?? – 2017-01-28
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    @DonAntonio I think I just fixed the typo. Is that better? – 2017-01-28
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    So $\;c\;$ doesn't even multiplies the first coordinate? – 2017-01-28
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    @DonAntonio That's all the textbook has written. I assume $(c)(a_1) = a_1$? – 2017-01-28
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    Your solution is incorrect, as $(0,0)\in V$ is the zero of the additive group $(V, +)$. Their solution isn't really well written, it should be "Let $a_1 \neq 0$ and $b_1$ such that $a_1 + b_1 \neq b_1$. Then $0×(a_1,0) + (b_1,0) = (a_1 + b_1, 0) \neq (b_1,0)$, so $0×(a_1,0)$ isn't the zero of the additive group $(V,+)$, so as defined, $(V, +, .)$ isn't a vector space." Or something like this (there must be nicer ways to write this) – 2017-01-28
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    @Max I understand now. Thank you. :) – 2017-01-28

2 Answers 2

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Your solution is not correct since $(0,0)$ belong to $V$. The problem lies on the scalar multiplication doesn't satisfies the properties:

$$\lambda (a,b)=(\lambda a, \lambda b)$$

$$0(a,b)=(0,0). $$

Infact your book say: $0(a_1,0)=(a_1,0)\neq (0,0)$ and thus the second property is not satisfied.

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    Thanks for the response. But how do you know that $(0, 0)$ belongs to $V$? We're not told what the field, $F$, is, so how can we assume that $a_1 = 0$ or $a_2 = 0$? – 2017-01-28
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    If $F$ is a field, $0\in F$, even if it is a trivial field. – 2017-01-28
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    That's interesting. Thank you very much for the assistance. – 2017-01-28
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    @ThePointer The set $\;V\;$ contains **all** the pairs $\;(a_1,a_2)\;$ , when $\;a_i\in F=$ a field, so zero *is* an element of $\;F\;$ and **thus** $\;(0,0)\in V\;$ . – 2017-01-28
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    @DonAntonio I see. I was unaware that $0$ is always an element of any field. Thanks for the help. :) – 2017-01-28
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    @ThePointer Yes, a field is special kind of ring, and thus an additive abelian group with an additional structure compatibile with the sum. Any group has the $0$ element, so $0$ belong to any field. – 2017-01-28
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The axioms of vector spaces require the identity element of the multiplication in $F$ to be the identity element of the scalar multiplication in $V$, but $$ (0,1) = 1\cdot (0,1) = (0,0) \neq (0,1) $$ Therefore $V$ cannot be a vector space.