Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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 $\mathbb{S} \neq \mathbb{T}$ –  user17762 Oct 18 '11 at 22:39
    
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 $\mathbb{R} \not\subseteq \mathbb{Q}$ –  kahen Oct 18 '11 at 22:40
    
@Sivaram: My mistake, fixed it. –  Javier Badia Oct 18 '11 at 22:41
    
@kahen: I probably should have mentioned that I'm working with real vector spaces here. I'll add that to the question. –  Javier Badia Oct 18 '11 at 22:42
add comment

2 Answers

up vote 5 down vote accepted

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 $\mathbb{T}$.

share|improve this answer
    
Thanks. I thought it was true, but wasn't sure. –  Javier Badia Oct 19 '11 at 1:40
add comment

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 $\mathbb{T}$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.