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I am given that S is a set, and V is a set such that $V = \{f:S→\mathbb R\}$ with $f_1, f_2, f \in V$ Show that $V$ is a vector space over $\mathbb R$ with scalar multiplication and addition

I am having troubles even getting started on this question. I have written my solution up for the addition part but I don't believe it is complete, let alone correct.

My thought process is that since $f_1(x) \in \mathbb R$ and $f_2(x) \in \mathbb R$ then $\implies$ $f_1(x) + f_2(x) \in \mathbb R$

Since $f_1(x) + f_2(x) = (f_1+f_2)(x)$ then that $\implies$ $(f_1 + f_2)(x) \in \mathbb R$ as well.

I am not fully understanding how to go about proving that it is closed under scalar multiplication and addition since there are no concrete numbers and I am confusing myself with theory.

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    $\begingroup$ what is a vector set over $\mathbb{R}$ ? you probably mean a vector space over $\mathbb{R}$ ? $\endgroup$ – KonKan Jan 18 '17 at 4:26
  • $\begingroup$ @KonKan I did mean vector space - I have edited it. Thank you $\endgroup$ – Torched90 Jan 18 '17 at 4:33
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Well, if $f_1,f_2\in V$, the addition on $V$ is probably defined by $(f_1+f_2)(x)=f_1(x)+f_2(x)$, hence $f_1+f_2$ is again a mapping from $S$ to $\mathbb R$, i.e., $V$ is closed under addition. You prove that it is closed under scalar multiplication in a similar fashion. You then need to show that all the other vector space axioms are satisfied, such as existence of zero, existence of additive inverse elements, etc. All of this is straightforward. The idea of this task is probably just to make you more familiar with the vector space axioms themselves.

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  • $\begingroup$ So is my above thought process correct. Should I just do a similar thing for the multiplication? $\endgroup$ – Torched90 Jan 18 '17 at 4:40
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    $\begingroup$ Yes, your process is correct, as long as you understand that $V$ is closed under addition and scalar multiplication, because these operations are defined by pointwise addition or multiplication of the function values. One could define these operations in a different way, i.e., not in the intuitive way that is used here, in which case $V$ may or may not be a vector space. $\endgroup$ – cthl Jan 18 '17 at 4:43

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