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I have a question in the test process of subspace.

To show that any particular subset is a subspace of any particular vector space, we generally check 'closed under addition', 'non-empty' and 'closed under scalar multiplication'.

However, this is exactly the same as proving that a set is a vector space. In the proof process, there is no process of confirming whether the subspace is a subspace of particular vector space.

Here is the example. Let's consider the problem "show $(x, 2x)$ is subspace of $\mathbb{R}^2$." In general, we check that it is a subspace as shown below.

  1. it is closed under scalar multiplication. $c(a, 2a)=(ca, 2ca)\in (x,2x)$
  2. it is closed under addition. $(a, 2a)+(b, 2b)=(a+b, 2(a+b))\in (x,2x)$
  3. Since it contains (0,0), it is non-empty.

However, there is no information of $\mathbb{R}^2$ space. If the given problem were to show that $(x, 2x)$ is a subspace of some other vector space encompassing $(x, 2x)$ but not $\mathbb{R}^2$, the proof process does not change.

To summarize the question, in general, in the process of proving that a subset is a subspace of a specific vector space, information about the vector space encompassing the subspace is omitted, so I think it is an incomplete proof. I hope you share your thoughts on this.

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The fact it's a subspace comes from the fact that you're checking addition using the addition rules from the encompassing vector space itself and the scalar multiplication from the field over which your initial vector space was defined. This is the information coming from $\mathbb{R}^2$. For instance, if you're trying to show that $\mathbb{R}$ is a vector subspace of $\mathbb{R}^2$, you're using addition in $\mathbb{R}^2$, so you'd show something like $(a,0)+(b,0)$ is included. So no, no information is omitted, it's just a bit hidden away. Effectively, if you have a vector space $(V,+_V,\cdot_V)$ and you're trying to show $W$ is a vector subspace, you show that $(W,+_V, \cdot_V)$ is a vector space. Notice where the $V$ appears here.

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