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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.2012-05-24
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    What does $k$ stand for?2012-05-24
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    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.2012-05-24
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    I understand its a constant, with $k\in R$2012-05-24
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    @fretty To be fair, $v_1$ and $v_2$ **are** defined -- they represent, respectively, the second and third component of a row vector.2012-05-24
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    @Soham Then the zero vector will be in your set if and only if $k$ has a specific value, which is...?2012-05-24
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    @fretty Oh yes, thanks, I did think about it for some time, but in the haze of multiple thoughts, I forgot to explore that bit. Yes, I am not actually told anything about $v_{1}$ hence was thinking how to go forth with it. I understand, the author wants us to just treat $v_{1}$ as that, that is associative under addition and follows scalar multiplication. I speculate, he wanted students to just explore the same two properties with $v_{2}, v_{3}$ but I was not sure. I thought ,was missing a trick here.But knowing what I know now, how do I go forth to find a basis for the same2012-05-24
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    @M Turgeon DId you mean $v_{2}$ and $v_{3}$ ? Nothing is mentioned anything about $v_{1}$2012-05-24
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    @M Turgeon, If indeed it is an element of the basis set, then for zero vector to be present in the vector space, I understand $v_{1}, v_{2},v_{3} $ has to be zero => k=02012-05-24
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    Ah right, I overlooked that bit. Yeah as people say, we have to have a "zero vector" in a vector space, in this paricular set of vectors this is $(0,0,0)$. Now you can use the condition to restrict to one possible value of $k$. Then for that value of $k$ check you get a vector space. As for a basis (when you DO get a vector space), can you find two vectors that span the space. Are these two vectors linearly independent?2012-05-24

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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.2012-05-24
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    But isnt dimension of $R^{n}$ generally n ?2012-05-24
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    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 1.2012-05-24
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    Oh thanks a ton!2012-05-24
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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.2012-05-24