Proving if a given set is a vector space
I am working from Erwin Kreyszig's book, where he mentions of a question.
Is the given set of vectors a vector space. If yes, determine the dimension and find a basis $(v_{1},v_{2},\cdots,v_{n} )$ denote components
All vectors in $\mathbb R^{3}$, such that $4v_{2}+v_{3}=k$
I have an intuitive understanding of vector space, as the output of a generator set (aka spanning set formed via basis set), is there a test for proving a particular set is a Vector Space?
Secondly, given this condition on $v_{2}$ and $v_{3}$ how do I go ahead to find the basis? I understand finding dimension from basis is a trivial exercise.
Help much appreciated
P.S: Apologies for formatting, but I am not able to render latex on my post. An additional link to some help would be much appreciated
$\endgroup$ 102 Answers
$\begingroup$Note that for a set to be a vector space it must be closed under addition and scalar multiplication, so that is what we need to check here. For addition, suppose $v=(v_1,v_2,v_3)$ and $w=(w_1,w_2,w_3)$ satisfy $4v_2+v_3=k$ and $4w_2+w_3=k$. Does $v+w$? Well, $4(v_2+w_2)+(v_3+w_3)=(4v_2+v_3)+(4w_2+w_3)=2k$, which is only equal to $k$ if $k=0$. So the set is only closed under addition if $k=0$. Similarly, if we let $r\in\mathbb R$ be a scalar then $4(rv_2)+rv_3=rk$ which is not necessarily equal to $k$ unless $k=0$. So the set you are given is only a vector space if $k=0$.
$\endgroup$ 1 $\begingroup$If $k$ is non-zero, then $v=(0,0,0)$ cannot satisfy $4v_2+v_3=k$, hence it is not a vector space. Otherwise, i.e. if $k=0$, you can verify that it is a vector space, since $$4u_2+u_3=0\text{ and }4v_2+v_3=0 \implies 4(u_2+v_2)+(u_3+v_3)=0$$ etc., etc. The basis is simply $\{(1,0,0),(0,-1,4)\}$.
$\endgroup$ 4
