I'm trying to prove that, for a linear transformation, it is worth that:
$f(a\bar{x}+b\bar{y})=af(\bar{x})+bf(\bar{y})$
Until now, I have proved that
$f(\bar{x}+\bar{y})=f(\bar{x})+f(\bar{y})$.
How can I use this result in order to prove that $f(a\bar{v}) = af(\bar{v})$ for every $a\in \mathbb{R} $ ?
The fact thar $f(a\bar{v})=af(\bar{v})$ for arbitrary $a$ is infact part of the defenition of linear operators. Nevertheless, for integer values of $a$ we have $$f(a\bar{v}) = f(\sum\bar{v})\text{(Sum it $a$ times)}$$ $$=\sum f(\bar{v})$$ $$=af(\bar{v})$$