From Friedberg....
" Let $V=\{(a_1 ,a_2):a_1 ,a_2\in\mathbb F\}$, where $\mathbb F$ is a field. Define addition of elements of $V$ coordinatewise, and for $c \in \mathbb F$ and $(a_1 , a_2) \in V$, define $$c(a_1 , a_2)=(a_1, 0).$$
Is $V$ a vector space over $\mathbb F$ with these operations? Justify your answer. "
My solution:
"" No, $V$ is not a vector space. Note that... $$c(a_1 , a_2)=(ca_1, ca_2)=(a_1 ,0)$$ ... implies...
$ca_1=a_1$ and $ca_2=0$
It follows that $c=1$ and $c=0$. ""
Where do I go from here? It seems that anywhere else it would've been enough to prove $c=1$ and $c=0$ (which is a contradiction), but my instinct tells me Linear Algebra requires more than this. Should I let $c=1$ and show that $1 \cdot a_2=a_2 \neq0$ or should I let $c=0$ and show that $0 \cdot a_1=0 \neq a_1?$ Neither? Both? I'm completely baffled. Any help would be greatly appreciated!
By stating that $c(a_1, a_2) = (ca_1,ca_2)$, you've made the assumption that scalar multiplication on the set $V$ has been defined in that way, but it hasn't! It's been defined exactly as stated in the question:
For all $c\in \mathbb{F}$ and $(a_1, a_2) \in V$, we define the scalar multiplication operation $\cdot$ by $$c\cdot(a_1,a_2) = (a_1,0).$$ We also define an addition $+$ by $$(a_1,a_2) + (b_1,b_2):=(a_1+a_2,b_1+b_2).$$
Holding that firmly in mind, you have to verify that the vector space axioms are satisfied by the vector space $V$ with the operations $+$ and $\cdot$. It's important to remember that the set $V$ by itself is not a vector space (over the field $\mathbb{F}$), and that the only thing that makes it a vector space is the structure you endow on it by defining the operations $+$ and $\cdot$. Once the problem defines $\cdot$ in a certain way, you forget everything you know about "traditional" scalar multiplication and work directly with the definition given. By introducing componentwise scalar multiplication into your proof in the way you have, you're saying that the operation defined ought to be equivalent to componentwise scalar multiplication, which isn't necessarily the case.
In this case, note that one of the axioms that needs to be satisfied in order for $(V, +, \cdot)$ to be a vector space is that for all $(a_1, a_2)\in V$, if $1\in\mathbb{F}$ is the multiplicative identity in $\mathbb{F}$, we should have: $$1\cdot(a_1,a_2) = (a_1,a_2).$$
In this case, you are actually correct that the multiplicative identity axiom does not hold. By definition, we have $$1\cdot(a_1, a_2) = (a_1, 0),$$ which does not equal $(a_1,a_2)$ unless $a_2 = 0$. A counterexample would be $1\cdot(1,1)=(1,0)\neq(1,1)$. This is enough to say that $(V,+,\cdot)$ is not a vector space, but you should also verify that distributivity over scalar addition does not hold either.