2
$\begingroup$

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.

  • 2
    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\}$2012-06-03
  • 0
    @Dylan, I have made the edit. Thankfully my moral compass is working alright these days ;)2012-06-03
  • 0
    @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.2012-06-03
  • 0
    @Soham Thanks. I thought this might be the case—usually the nefarious ones don't make themselves so easily identifiable. [You'd be surprised, though.] Your question is, of course, completely appropriate for this site.2012-06-03
  • 2
    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.2012-06-03
  • 0
    @Robert As it is a subspace, the dimension of this need not be equal to 4, as is evident that its dimension is 2. That is my understanding, correct me if I am wrong. Regarding the second bit, you mean to say there are C(4,2) combinations in which 2 of the variables are made zero and the basis found out? Hmm, if so, I think you have a point there.2012-06-03
  • 0
    @Dylan ;) $[]$ $[]$2012-06-03
  • 1
    @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.2012-06-03
  • 0
    If I asked you for one week of the year and you said "June 12th" then you would have given me one day of one week. The problem here is similar, a vector space has many possible bases and you have to give one of them...you have given one vector in a basis.2012-06-06
  • 0
    @RobertIsrael Yes, I get the point, thanks.2012-06-10
  • 0
    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

1 Answers 1

1

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.