I'm doing an exercise that goes like this:
Determine whether $\mathbb{S}$ and $ \mathbb{T}$ are equal:
$\mathbb{S} = \langle(1, 0, 2), (1, 1, -1)\rangle$ $\mathbb{T} = \{x \in \mathbb{R}^3 \mid 2x_1 - 3x_2 - x_3 = 0\}$
The way I did it is first I check whether the dimensions are the same. If they are, I need to check whether one is a subspace of another. I take a "sample" vector from $\mathbb{S}$: $s \in \mathbb{S} = a(1, 0, 2) + b(1, 1, -1) = (a+b, b, 2a-b)$ with $a, b \in \mathbb{R}$. I then check whether they satisfy $\mathbb{T}$'s equation: $2(a+b) - 3b - 2a-b = 0$, which is true no matter what $a$ and $b$ are. Therefore, the two vector spaces are equal.
While doing a few more problems like this, I noticed that I could just take the vectors of $\mathbb{S}$'s basis and see if they belong in $\mathbb{T}$, and if they do then $\mathbb{S} \subseteq \mathbb{T}$. While this sound right and it seems to make sense, I'm not sure. Is this always true?