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I'm a student who is trying to learn Linear Algebra. In the textbook I'm using, determining subspaces is only shown with examples of functions. However some of the exercises ask whether or not vectors are subspaces. Are my following assumptions correct?

1) All vectors in $ℝ^5$ of the form $(a, b, c, d , e)$ where two entries are zero.

  • For this, it is closed under addition. $0+0=0$ for any two entries are $0$.
  • Closed under scalar mult. since zero times anything will still be zero.

2) All vectors in $ℝ^6$ of the form $(a, b, c, d , e, f)$ where all the entries are integers.

  • Not closed under addition, because adding a negative integer would turn the result from pos. to neg. and vice versa

  • Not closed under scalar mult. since multiplying by neg. integer would yield a pos. integer negative and vice versa

3) All vectors in $ℝ^6$ of the form $(a, b, c, d , e, f)$ where $a+c+e = b+d+f$

  • Not sure how to approach this. I guess intuitively I would say that yes, it is closed under addition and scalar mult, but I don't know how to prove it.
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    $\begingroup$ For point 1) No, you aren't right, it's not closed for addition : counterexample $(0,0,1,1,1)+(1,1,1,0,0)=(1,1,2,1,1)$ $\endgroup$
    – Jean Marie
    Commented Feb 4, 2018 at 23:29
  • $\begingroup$ For number 2), integer vectors are closed under addition: The sum of two integers is always an integer, and the positive/negative distinction doesn't matter. The reason it's not closed under scalar multiplication has nothing to do with positive/negative integers, either. Try multiplication by the scalar $\frac12$. $\endgroup$ Commented Feb 4, 2018 at 23:34
  • $\begingroup$ Please remember that you can choose an aswer among the given is the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$
    – user
    Commented Mar 9, 2018 at 22:34

2 Answers 2

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1) All vectors in $ℝ^5$ of the form $(a, b, c, d , e)$ where two entries are zero.

  • if the request is exactly 2 vectors it is not a subspace since it does not contain the zero vector
  • anyway it is not a subspace since we can easily find couterexamples, eg $(0,0,1,1,1)+(1,1,1,0,0)=(1,1,2,1,1)$

2) All vectors in $ℝ^6$ of the form $(a, b, c, d , e, f)$ where all the entries are integers.

  • note that it is closed under addition but it is not closed under scalar multiplication thus it is not a subspace

3) All vectors in $ℝ^6$ of the form $(a, b, c, d , e, f)$ where $a+c+e = b+d+f$

  • let verify by definition

    a) $0$ is in the subset

    b) $kv_1+hv_2$ is in the subset

thus it is a subspace.

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  • $\begingroup$ In 1), we're not told exactly two entries are zero. How would you show it's not a subspace under the interpretation that "two entries" means "at least two entries"? $\endgroup$ Commented Feb 4, 2018 at 23:35
  • $\begingroup$ The OP's solution to number 2) is far from "correct" $\endgroup$ Commented Feb 4, 2018 at 23:36
  • $\begingroup$ @GTonyJacobs ok I've interpreted the OP in another ways, I fix, Thanks! $\endgroup$
    – user
    Commented Feb 4, 2018 at 23:36
  • $\begingroup$ @GTonyJacobs for #2 I meant it was correct the conclusion, I've fixed, is it ok now? Thanks $\endgroup$
    – user
    Commented Feb 4, 2018 at 23:42
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1) Consider the vectors $(1,2,3,0,0)$ and $(1,0,0,4,5)$ which are both elements of the space. Their sum has no entries equal to 0 so this is not closed under addition

2) It does not matter if the sign changes from positive to negative. If all entries are integers, their sum will also be an integer (maybe positive and maybe negative). The real problem here is that multiplying by something like $0.5$ would result in a vector outside of the space.

3)

  • If we have the vector $(a,b,c,d,e,f)$ where $a+b+c=d+e+f$ (i.e. the vector is in the vector space) and we multiply by $\alpha \in \mathbf{R}$ then the resulting vector is $$(\alpha \cdot a,\alpha \cdot b,\alpha \cdot c,\alpha \cdot d,\alpha \cdot e,\alpha \cdot f)$$. Checking the sum of the first 3 components and the last three components we see $$\alpha \cdot a + \alpha \cdot b + \alpha \cdot c = \alpha \cdot (a+b+c) = \alpha \cdot (d+e+f) = \alpha \cdot d + \alpha \cdot e + \alpha \cdot f $$ So it is indeed closed under scalar multiplication
  • Consider the vector $(x_1,x_2,x_3,x_4,x_5,x_6)$ and $(y_1,y_2,y_3,y_4,y_5,y_6)$ such that the sum of the first three and the sum of the last three components of each vector are equal. The addition of these two vectors gives $$(x_1 + y_1,x_2+ y_2,x_3+y_3,x_4+y_4,x_5+y_5,x_6+y_6)$$ We then check the sum of the first three components of this vector and see if it equals the sum of the last three \begin{align} (x_1 + y_1) + (x_2 + y_2) + (x_3+y_3) &= (x_1+x_2+x_3)+(y_1+y_2+y_3) \\ &= (x_4+x_5+x_6) + (y_4+y_5+y_6)\\ & = (x_4 + y_4) + (x_5 + y_5) + (x_6+y_6) \end{align} Where the second line is by definition of being in the vector space. So this is closed under addition since the sum of the first three components is equal to the sum of the last three components
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