Determining which subsets of real numbers are subspaces 
Determine which of the following are subspaces of $\Bbb{R}^n ?$  
(a) $\{(a_1,a_2,...,a_n) \in \Bbb{R}^n: a_1 = 0\}$.
(b) $\{(a_1,a_2,...,a_n) \in \Bbb{R}^n: a_1 = 3\}$.  
(c) $\{(a_1,a_2,...,a_n) \in \Bbb{R}^n: a_1+a_2+...+a_n =1\}$
(d) $\{(a_1,a_2,...,a_n) \in \Bbb{R}^n: a_1^2 = a_2\}$.  
(e) $\{(a_1,a_2,...,a_n) \in \Bbb{R}^n: 2a_1+3a_2 =0\}$.  

(a) Does seem to be a subspace since it contains the $0$ vector and satisfies closure under scalar multiplication and vector addition.
(b) Isn't a subspace since it fails scalar multiplication since you can multiply by $-1$ and $a_1$ won't equal $3$. 
(c) Also isnt a subspace since it can't contain the $0$ vector and be $= 1$ and wouldn't satisfy closure under addition since you can have something like $(0,1,0)$ and $(1,0,0)$. 
(d) Satisfies $0$ vector, it should pass closure under addition, I'm not sure about multiplication though. If we had the vector $a=(a_1,a_2)$ and multiplied it by a negative $x$   so it became $-x(a^2)$ it should still equal $a_2$ but I'm unsure.
(e) Satisfies $0$ vector, addition works out; you can write it as $a = (a_1,a_2) b = (b_1, b_2)$ and $(a_1+b_1, a_2+b_2)$ then $2(a_1+b_1) + 3(a_2+b_2) =0$ so $a+b$ is in the set. Finally if we had an $x(2(a_1)+3(a_2))=0$ then $x \cdot 0=0$ so this should be a subspace. 
How do these look? I know the explanations aren't thorough but I'm trying to see if I have the main ideas correct. Thank you. 
 A: 
How do these look, I know the explanations aren't thorough but I'm trying to see if I have the main ideas correct. Thank you

It seems you understood the general idea. You are right about (a), (b), (c) and (e).

Yea apologizes about the formatting, I'll edit the italics its just supposed to be a1 squared. Technically, the original problem was a2 squared = a1 but I presumed it wouldn't matter if I switched them to make writing it easier.

For (d): note that $(1,1,\ldots)$ would satisfy $a_1^2=a_2$ (or $a_1=a_2^2$) but $2\cdot(1,1,\ldots)=(2,2,\ldots)$?
A: Your general procedure is right: check for the zero vector, check closure under addition, check closure under scalar multiplication. You seem to have done a good job on (a), (b), (c), and (e) ((b) also failed zero vector, you could have used that too).
For (d), you haven't convinced me that it's closed under addition, and you haven't convinced yourself of scalar multiplication. Try some easy concrete examples, like $(1,1)$ or $(-1,1)$, to see if you can get a handle on this.
