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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?

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  • $\begingroup$ If I understand correctly, you should have $x \in \mathbb{R}^3$ in the definition of $\mathbb{T}$. If $x \in \mathbb{R}^4$, you just have $\mathbb{S} \subset \mathbb{T}$ but $\mathbb{S} \neq \mathbb{T}$ $\endgroup$
    – user17762
    Oct 18, 2011 at 22:39
  • $\begingroup$ As stated in the title, the answer is no — provided we allow the vector spaces to be over different fields: $1$ is a basis vector for $\mathbb{R}$ (as an $\mathbb{R}$-vector space) and $\{1\} \subseteq \mathbb{Q}$ (considered as a $\mathbb{Q}$-vector space), but $\mathbb{R} \not\subseteq \mathbb{Q}$ $\endgroup$
    – kahen
    Oct 18, 2011 at 22:40
  • $\begingroup$ @Sivaram: My mistake, fixed it. $\endgroup$
    – Javier
    Oct 18, 2011 at 22:41
  • $\begingroup$ @kahen: I probably should have mentioned that I'm working with real vector spaces here. I'll add that to the question. $\endgroup$
    – Javier
    Oct 18, 2011 at 22:42

2 Answers 2

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Every vector in $\mathbb{S}$ is a linear combination of the basis vectors and $\mathbb{T}$ is closed under linear combinations. It $\mathbb{T}$ contains a basis for $\mathbb{S}$ then every vector in $\mathbb{S}$ is an element of $\mathbb{T}$.

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  • $\begingroup$ Thanks. I thought it was true, but wasn't sure. $\endgroup$
    – Javier
    Oct 19, 2011 at 1:40
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If $\mathbf{W}$ is a subspace of $\mathbf{V}$, and $T$ is a subset of $\mathbf{V}$, then $$T\subseteq \mathbf{W}\text{ implies }\mathrm{span}(T)\leq \mathbf{W}.$$

This follows because $\mathrm{span}(T)$ is the smallest subspace of $\mathbf{V}$ that contains $T$. In particular, it must be contained in the subspace that contains $T$.

In particular, if $T$ is a basis for a subspace $\mathbf{Z}$, then $T\subseteq \mathbf{W}$ implies $\mathbf{Z}\leq \mathbf{W}$.

Since you are given $\mathbb{S}$ as the span of its basis, to check whether $\mathbb{S}\subseteq\mathbb{T}$, it suffices to check if the given basis (or any spanning set for $\mathbb{S}$) is contained in $\mathbb{T}$.

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