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Is the given set of vectors a vector space? Give reasons. If your answer is yes, determine the dimension and find a basis. denote components.)

All vectors in $\mathbb{R^3}$ with $v_1-v_2+2v_3=0$.

All vectors in $\mathbb{R^3}$ with $3v_2+v_3=k$.

All vectors in $\mathbb{R^2}$ with $v_1 \geqq v_2$.

All vectors in $\mathbb{R^n}$ with the first $n-2$ components zero.

Being new to vector space, I really don't understand what's going on. Please help. Can anyone explain to me?

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    Start by using the site's search feature to search for obvious things like "given set vector space," and you will hit dozens of questions like this that will help you. If you do still need to ask a question, type it out (you can ask for help formatting if that's the problem). – 2017-01-21
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    Do you know what a vector space is? As in, the axioms? If you did, you would know to start checking the sets against them. – 2017-01-21
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    And yes, consider sharing your ideas and thought process for the given problem or atleast how did you proceeded. – 2017-01-21
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    You must check that if for some of these sets the axioms that define a vector space holds. Then start memorizing the axioms that define a vector space. – 2017-01-21
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    Please replace the image of the text with the text itself. – 2017-01-21

1 Answers 1

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You've been given, for each case, a set of vectors. For example, the first set is specified as

All vectors in $R^3$ with $v_1-v_2+2v_3=0$.

In set builder notation, this would be $$\{v\in R^3|v_1 - v_2 + 2 v_3 = 0\}$$ Note that this specifies a subset of the vector space $R^3$ (the use of the term "vectors" in the description already tells that; I very much suspect that $R$ is meant to be the set of real numbers, more often written as $\mathbb R$; however that's irrelevant, as it works the same over every vector space over a field containing the integers; although in the third task you need at least an ordered field).

Now your task is to check whether this set, equipped with the corresponding subset restrictions of the vector space operations of $R^3$, fulfills the axioms of a vector space.

Since you know that the set is a subset of a vector space, some of the properties, like associativity, are a given. Indeed, you only need to check two things:

  • For every two vectors $v,w$ in that set, $v+w$ is in the set.
  • For every $v$ in that set and every $\alpha$ in the underlying field (probably $\mathbb R$) also $\alpha v$ is in the set.

All other properties of a vector space can be derived from those two conditions and the fact that $R^3$ is a vector space.

For this first case, it is indeed the case:

  • $(v+w)_1 - (v+w)_2 + 2(v+w)_3 = (v_1+w_1) - (v_2+w_2) + 2(v_3+w_3) = (v_1 - v_2 + 2v_3) + (w_1 - w_2 + 2w_3) = 0 + 0 = 0$ $\implies$ the equation is fulfilled for the sum.

  • $(\alpha v)_1 - (\alpha v)_2 + 2(\alpha v)_3 = \alpha v_1 - \alpha v_2 + 2\alpha v_3 = \alpha(v_1 - v_2 + 2v_3) = \alpha 0 = 0$ $\implies$ the equation is fulfilled for the multiple.

With this, you should now be able to do it for the other cases.

Here are some hints:

  • For case 2, it's easiest to directly check the zero vector.

  • For case 3, consider especially $\alpha<0$.