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A set S consists of those vectors with a finite number of nonzero components.

Given a vector space V with infinite dimensions, or $V = R^\infty$, I am trying to prove that the set S is a subspace of it.

This is Problem 18 in Section 7.1 (Vector Spaces and Subspaces), page 285, from Linear Algebra by Jeffrey Holt.


By definition, to be a subspace, S must:

  1. Contain the zero vector
  2. Be closed under vector addition
  3. Be closed under scalar multiplication

I immediately want to say that this proof is false because of the first point above with the zero vector. Since S consists of those vectors with a finite number of nonzero components, it makes me think that it cannot contain the zero vector which would have the same number of components as S, with all of them being 0 (e.g. The zero vector contains a finite number of zero components!)

Can someone help me understand this better? Is my logic correct?

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  • $\begingroup$ $R^\infty$ is not a good notation, nor a legit one, AFAIK. $\endgroup$
    – user228113
    Commented Nov 2, 2016 at 22:00
  • $\begingroup$ And, what do you mean by component? $\endgroup$
    – user228113
    Commented Nov 2, 2016 at 22:04
  • $\begingroup$ What does it mean for the zero vector to "have the same number of components as $S$?" Can you give some examples of elements in $S$ and explain why the zero vector does not fit this description? $\endgroup$
    – Matt
    Commented Nov 2, 2016 at 22:05
  • $\begingroup$ Sorry, that is how my textbook wrote the question. $\endgroup$
    – zhughes3
    Commented Nov 2, 2016 at 22:05
  • $\begingroup$ @G. Sassatelli Matt I am trying to show that S is a subspace, which means it must contain the zero vector. For it to be a member of S, it must have the same number of components as a vector in S. So if S is in $R^3$, a valid vector would <1, 2, 3> $\endgroup$
    – zhughes3
    Commented Nov 2, 2016 at 22:07

1 Answer 1

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Your headline question is poorly phrased. Your longer description is better, though please note the comments about notation.

$S$ is a subspace. Let's look at the 3 rules.

  1. $0 \in S$. Of course it is, it has a finite number of nonzero components, i.e. $0$ of them!
  2. $s,t \in S \implies s+t \in S$. The indices of the nonzero components of $s+t$ is at most the union of the indices for $s$ and $t$, hence finite.
  3. $s \in S \implies \lambda s \in S$ for any scalar $\lambda$. Becuase we have already covered off the zero vector, we may assume $\lambda \neq 0$. Then the zero components of $s$ and $\lambda s$ are identical.
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  • $\begingroup$ In this particular context, can I interpret "s + t is at most the union of the indices for s and t" as "s + t is at most the sum of the indices for s and t"? $\endgroup$
    – zhughes3
    Commented Nov 3, 2016 at 1:44
  • $\begingroup$ Not really. I should have been more precise. Let $s_i$ be the $i$-th component of s. $i$ is the index. For $(s+t)_i$ to be nonzero, at least one of $s_i$ and $t_i$ must be nonzero. This can only occur a finite number of times. $\endgroup$
    – user383778
    Commented Nov 5, 2016 at 2:09

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