1
$\begingroup$

I understand that in order to have a linear space I should sattisfy both closeness of the sum and the multiplication by a scalar ( and they must satisfy certain properties). However I have trouble in understanding what the following statement asks me:

(Problem 2.from the book in section 1.)

  1. Which of the following sets of vectors $$ x=(x_{1},x_{2},...,x_{n})$$ in the space $L_{n}$ are linear spaces?

$$a) The\space set\space such\space that \space x_{1}+x_{2}+...+x_{n} = 0$$ $$b) The \space set\space such\space that \space x_{1}+x_{2}+...+x_{n} = 1$$

Am I right to think about that sum as the action of sliding the number line and giving me the sum which converges to a single point where as the second statement tells me that it doesn't converges but instead it falls behind by the unit element?

And how is a) a linear space and b) isn't?

  • 0
    Hint: Test your definition of "linear space", which probably involves closure under addition, on some sort of generic sample of each type. – 2017-02-06
  • 0
    In order for a space to be linear, there are 8 axioms that need be satisfied and they are described on this [wikipedia site](https://en.wikipedia.org/wiki/Vector_space#Definition). Does the first set satisfy all of the axioms? Regarding the second set. A linear space **always** includes the **zero vector.** Can this happen in the second case? – 2017-02-06
  • 0
    By the way do you mean linear *subspaces*? – 2017-02-06
  • 0
    Mmmm no, just linear space... and If I try to do closure under addition I think it works for the first case but not for the second. However I'm not sure if that's the correct approach. – 2017-02-06

0 Answers 0