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.
Best Answer
It seems you understood the general idea. You are right about (a), (b), (c) and (e).
For (d): note that $(1,1,\ldots)$ would satisfy $a_1^2=a_2$ (or $a_1=a_2^2$) but $2\cdot(1,1,\ldots)=(2,2,\ldots)$?