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I am given the vector space of $\mathbb{R}^2$ and one of its element which is $(5,7)$. How can I determine the "smallest" subspace $A$ of $\mathbb{R}^2$, that contains $(5,7)$?

average_discrete_math_enjoyer
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    Do you know what subspaces of $\mathbb{R}^2$ look like ? – TheSilverDoe Nov 04 '20 at 13:13
  • Yes, I have been taught what a subspace of $\mathbb{R}^2$ is. Specifically, we have learned that for a set A to be subspace of $\mathbb{R}^2$, it must contain the $\vec{0}_A \in \mathbb{R}^2$, and we must be able to add and multiply its elements, so that these elementss are again in $\mathbb{R}^2$. I am confused as of what does the element (5,7) really show us? And how can I find a subspace of a single element. – average_discrete_math_enjoyer Nov 04 '20 at 13:16
  • I suppose your exercise deals with $\mathbb{R}$-vector spaces, right? It is also interesting to think about the smallest $\mathbb{Q}$-vector space containing your point. – Aurelio Nov 04 '20 at 13:17
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    @average_discrete_math_enjoyer I mean, can you give me examples of subspaces of $\mathbb{R}^2$ ? – TheSilverDoe Nov 04 '20 at 13:18
  • @average_discrete_math_enjoyer, you are not looking for a subspace of a single element, but a subspace containing at least your given element $(5,7)$. Call $U$ this subspace, and start by thinking which other elements you must necessarily include in $U$. For example, you cannot avoid having $0\in U$. – Aurelio Nov 04 '20 at 13:19
  • @TheSilverDoe Yes, e.g. a set $A = { (x,y) : 5x +3y=0 }$ is a subspace of $\mathbb{R}^2$. – average_discrete_math_enjoyer Nov 04 '20 at 13:20
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    @average_discrete_math_enjoyer Good answer ! And geometrically, can you describe what are the subspaces of $\mathbb{R}^2$ ? I mean, can you picture them ? How they look like ? – TheSilverDoe Nov 04 '20 at 13:21
  • It is worth pointing out the generalization to the problem... the smallest subspace containing each of the $k$ vectors $v_1,v_2,\dots,v_k$ is equal to the span of those vectors, $\text{span}(v_1,v_2,\dots,v_k) = {c_1v_1+c_2v_2+\dots+c_kv_k~:~c_i\in\Bbb F~\forall i}$ where $\Bbb F$ is your scalar field, whatever it happened to be. The dimension of that subspace will be at most $k$ (it could be $k$, it could be less, it depends on whether the vectors $v_1,v_2,\dots,v_k$ are linearly independent or not and how much redundancy there is if they are dependent). – JMoravitz Nov 04 '20 at 13:26
  • @TheSilverDoe Our teacher didn't go into much detail on that. I would assume that this line which goes through the origin, can produce vector spaces in $\mathbb{R}^2$, for which the definition of subspaces is true? As you have understood thus far our teacher is pretty useless, so I am trying to understand all of these concepts way faster that I think I can comprehend. – average_discrete_math_enjoyer Nov 04 '20 at 13:28
  • What SilverDoe is alluding to is that all subspaces of $\Bbb R^n$ (over scalar field $\Bbb R$) will be the origin by itself, a line passing through the origin, planes passing through the origin, or hyperplanes (the higher-dimensional analogues to planes) passing through the origin. – JMoravitz Nov 04 '20 at 13:30
  • Does this answer your question? Span of a subset of a vector space is the smallest subspace containing that set – JMoravitz Nov 04 '20 at 13:34
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    @average_discrete_math_enjoyer You are right about the lines. Here is a way to understand completely the subspaces of $\mathbb{R}^2$. First a subspace must contain $0$, by definition. You can check that $\lbrace 0 \rbrace$ is a subspace of $\mathbb{R}^2$, this is the first (trivial) example. Now let's suppose that you have another point $x \in \mathbb{R}^2$ in your subspace. Then you can see, by definition, that the whole line through $0$ and $x$ must be included in your subspace. Conversely, you can check that such a line defines a subspace. So you have also lines. (...) – TheSilverDoe Nov 04 '20 at 13:35
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    @average_discrete_math_enjoyer (...) Finally, let's suppose that you have a line and another point outside the line, inside your subspace. Then you can see that the whole plane $\mathbb{R}^2$ must be cointained in your subspace, therefore the subspace is $\mathbb{R}^2$. So to conclude, the subspaces of $\mathbb{R}^2$ are exactly : $\lbrace 0 \rbrace$, the lines passing through $0$, and $\mathbb{R}^2$. And nothing else. – TheSilverDoe Nov 04 '20 at 13:35
  • @JMoravitz I am afraid that your comment is misleading, due to a terminology issue. Commonly, a hyperplane is a codimension-$1$ subspace. If you want to list families of subspaces of $\mathbb {R}^n$, you should include also all $k$-planes for $2\leq k\leq n-2$. – Aurelio Nov 04 '20 at 13:35

1 Answers1

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The smallest subspace which contains $(5,7)$ is one-dimensional:

$$\{t(5,7): t \in \mathbb R\}.$$

Fred
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  • Why the down votes?? There is literally no better answer! – PierreCarre Nov 04 '20 at 13:21
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    @PierreCarre This is maybe the best answer, but it is also the less pedagogical one. – TheSilverDoe Nov 04 '20 at 13:22
  • @TheSilverDoe That's one way to see it. But, taking a look at this answer, the OP may guess exactly what you were trying transmit when you asked "How they look like?". Pedagogy can take many forms. – PierreCarre Nov 04 '20 at 13:26
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    Would you mind explaining the idea behind reaching this result? – average_discrete_math_enjoyer Nov 04 '20 at 13:31
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    @average_discrete_math_enjoyer If $(5,7) \in A$, since it must be closed with respect to the multiplication by scalars, it must also contain all multiples of $(5,7)$. Since this already forms a subspace, you don't to add anything else. – PierreCarre Nov 04 '20 at 13:34