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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?

Soham

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 suggest you have a look at each question you have asked and, if you are happy with one or more of the answers, then accept one of the answers that makes you happy.2012-06-11

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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.