0
$\begingroup$

I am asking this question to clarify my understanding of set of functions as vector spaces. For instance, if i have the function $$f:[a,b]->[c,d]$$ Where a,b,c,d ∈ R Then $$f(s)∈[c,d]$$ For any s∈R with the definition of scalar multiplication, let r∈R $$(r*f)(s) = r * (f (s)) ∈[r*c, r*d]$$ Since the codomain of the multiplication can be outside of [c,d] when r>1, we can conclude that the set of $f$ is not a vector space since it is not closed under scalar multiplication. With this result it becomes quite obvious that any set of functions that have finite codomains are not closed under scalar multiplication. Is this correct? I still feel a bit confused about sets of functions being vector spaces, I just started learning linear algebra.

$\endgroup$
2
  • $\begingroup$ Can you adjust the title, it seems you mean not finite codomain, but non-zero finite length interval codomain. $\endgroup$
    – mvw
    Jan 6, 2016 at 23:04
  • $\begingroup$ Usually when you consider function spaces as vector spaces you use (pointwise) the structure on the codomain. Functions from any set to $\mathbb{R}$ form a vector space since $\mathbb{R}$ is. In this case, $[c,d]$ isn't a vector space. $\endgroup$
    – Abel
    Jan 6, 2016 at 23:38

2 Answers 2

1
$\begingroup$

What you said is true, but you can also see it if you look at addition of functions for example define some functions $f(x)=d$ and $g(x)=c$, where $x \in [a,b]$, these are constant function and it is obvious that it holds $$(f+g)(x)=f(x)+g(x)=c+d.$$ Which in case of c and d are of same sign implies that set of such function is not closed for addition and can't be vector space for these operations (additions and multiplying with scalar).

$\endgroup$
0
$\begingroup$

$$ V = \{ f : \{ 0 \} \to \{ 0 \} \mid f(x) = 0 \} = \{ 0 \} $$ seems to work as a vector space, even with the codomain written as $[0,0]$.

$\endgroup$
6
  • $\begingroup$ That is differently defined set. For set defined in question with a segment as a codomain it doesn't work. $\endgroup$
    – Lale221
    Jan 6, 2016 at 22:50
  • $\begingroup$ Your title asks for a finite codomain. $\endgroup$
    – mvw
    Jan 6, 2016 at 22:51
  • 1
    $\begingroup$ You might view $\{0\}$ as $[0,0]$. $\endgroup$ Jan 6, 2016 at 23:06
  • $\begingroup$ @mvw not mine and it is misleading but in question he asks for $x \in [c,d]$ and while you are correct (because he didn't exclude c=d) I don't think he wanted a special case that works but rather an explanation why it doesn't work in general. $\endgroup$
    – Lale221
    Jan 6, 2016 at 23:06
  • $\begingroup$ That's why we try to sharpen the question. $\endgroup$
    – mvw
    Jan 6, 2016 at 23:08

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .