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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

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You can render LaTeX by enclosing it with dollar signs. I have edited your post accordingly. – Alex Becker May 24 '12 at 17:10
Thanks a ton. Help much appreciated – Soham May 24 '12 at 17:10
What does $k$ stand for? – M Turgeon May 24 '12 at 17:11
A vector space is a set of things that make an abelian group under addition and have a scalar multiplication with distributivity properties (scalars being taken from some field). See wikipedia for the axioms. Check these proprties and you have a vector space. As for a basis of your given space you havent defined what v_1, v_2, k are. – fretty May 24 '12 at 17:12
I understand its a constant, with $k\in R$ – Soham May 24 '12 at 17:13
up vote 1 down vote accepted

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 $\{(1,0,0),(0,-1,4)\}$.

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Mucho thanks, much appreciated. – Soham May 24 '12 at 17:26
But isnt dimension of $R^{n}$ generally n ? – Soham May 24 '12 at 17:27
Yes, the dimension of $\mathbb{R}^n$ is $n$ (as a vector space over $\mathbb{R}$) but the space you are considering here is inside $\mathbb{R}^3$, not equal to it. When you draw lines in the plane you aren't drawing 2 dimensional things but might be describing a subspace of dimension 1. – fretty May 24 '12 at 17:35
Oh thanks a ton! – Soham May 24 '12 at 17:42

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 $k=0$.

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Awesome, logic! Mucho thanks. But I am going to mark up Somabha's answer, primarily because it answered every question of mine. – Soham May 24 '12 at 17:25

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