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I am putting an exam question of mine, which I hope complies with the overall policy of this forum. I have attempted it myself and I am checking effectively if my answer is right or not.

EDIT: This question paper is already turned in and done with. Nothing can change my answer anymore. I am just trying to clarify my concepts by bouncing my thoughts with others on this forum so as to understand how has my concepts crystallized. The exam is a public examination and hence, no solution or discussion will be held (naturally!)

Prove that the set $V$ of vectors $\{x_{1},x_{2},x_{3},x_{4}\}$ which satisfy the equation$x_{1}+x_{2}+2x_{3}+x_{4}=0$ and $2x_{1}+3x_{2}-x_{3}+x_{4}=0$ is a subspace of $\mathbb R^{4}$. What is the dimension and find one of its bases?


My approach: To prove to be a subspace, it will have to be proved that its a vector space with null vector in its set. if $v$ and $u$ are solutions, then $v+u$ is also solution, since $0+0 =0$ and naturally $cv=0$. To find its dimension, I wrote it as a matrix with $2$ rows as zero rows and $2$ rows as non zero rows, and since one non zero row is not a scaled version of the other, hence rank $=2$ , hence to find the dimension of this matrix, where X is the null space of this matrix, we have to find the nullity , therefore nullity = number of unknowns - rank = $4-2 =2$, so we can let two of the variables take any value, and still solve the system of eqns because $\dim =2.$ Hence taking $(x_{3},x_{4})=\{1,0\}$, we find $\{-7,5,1,0\}$ is a base.

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I see no big flaw in your argument, and your answer is correct, apart from the fact that a two dimensional vector space needs two vectors as base, not just $\{-7, 5, 1, 0\}$, but also let's say $\{-2, 1, 0, 1\}$ – Arthur Jun 3 '12 at 21:19
@Dylan, I have made the edit. Thankfully my moral compass is working alright these days ;) – Soham Jun 3 '12 at 21:21
@Arthur, thanks for the clarification, yes of course there is another, and since they have asked only one of the bases so gave only one. Do make it as an answer so that I can tick it as so. – Soham Jun 3 '12 at 21:21
A basis of an $n$-dimensional space consists of $n$ vectors. They asked for one basis, not for one basis vector. So your answer is incomplete. – Robert Israel Jun 3 '12 at 21:35
@Soham: I didn't say $4$. You are correct that the dimension is $2$. All I am saying is that you need two vectors, not just one, to make a basis of this space. – Robert Israel Jun 3 '12 at 22:12
up vote 1 down vote accepted

Your answer is correct, apart from the fact that a two-dimensional vector space needs two vectors as base, not just $\{−7,5,1,0\}$, but also let's say $\{−2,1,0,1\}$.

Remember that the two basis vectors together is one basis for a $2$-dimensional vector space.

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