Why is the "non-zero vector" hypothesis necessary here? I'm self-studying linear algebra and don't want to post too many details here in case someone is using the same textbook. I ran across this simple exercise where I think there is an extra hypothesis.
To prove:
Suppose $ A \in \mathbb{R}^n, A \neq 0, c \in \mathbb{R}$.
Then the set $ E = \{ x : A \cdot x \ge c,  x \in \mathbb{R}^n \} $ is convex.
My Attempt:
We need to show that for all $ 0 \le \alpha \le 1 $ and $x, y \in E $ that
$ w = \alpha x + (1-\alpha) y \in E.$
Consider $ A \cdot w = A \cdot (\alpha x + (1 - \alpha) y) = \alpha (A \cdot x) + (1 - \alpha)(A \cdot y)$ by the linearity of the dot product. This last expression is at least $c$ since $A \cdot x \ge c$ and $A \cdot y \ge c$, and the expression is the weighted average of these two real numbers. Hence $w \in E$ as desired.
My Question:
Why is the $A \ne 0$ hypothesis necessary? Suppose $A = 0$. If $ c \leq 0$, then $E = \mathbb{R}^n$, and the entire vector space is convex since it is closed under addition and scalar multiplication. If $ c \gt 0 $, then $E = \varnothing$ and is "vacuously" convex.
 A: The hypothesis isn't necessary for the reasons you mentioned yourself. Besides what copper.hat says in their comment above, another reason one might decide to enforce $A \neq 0$ is that  the nature of the space $E$ changes if $A=0$ is allowed. As long as $A\neq 0$, you may understand $E$ as  a "closed half space", something akin to the set of $(x,y,z)$ in Euclidean space with $z\geq 0$. So, one can interpret this exercise as "prove closed half-spaces are convex". There is a good reason to care about closed half spaces in this context because actually closed convex sets are exactly the sets which can be expressed as intersections of closed half spaces. This is basically the supporting hyperplane theorem.
By the way, while I would typically say "more words less symbols" is good style, I think in this case writing
$$\alpha (A \cdot x) + (1 - \alpha)(A \cdot y) \geq \alpha c + (1-\alpha)c = c$$
might be more compelling than what you wrote about "weighted averages". Just my opinion though, everything basically looks good to me.
