# Thread: Dimensions of linear subspaces

1. ## Dimensions of linear subspaces

i) 2 Dimensions

(1,0,0,0) * 8 = (8,0,0,0)
(1,0,0,0) * 15=(15,0,0,0)

ii) 1 Dimension because I have one vector only? Abb (N,R) means f:N-->R

iii) Dont know. How do I start here? I have modulo 3, but thats all. How do I find out the dimension here?

iv) 2 Dimensions
(1,1,0) + (0,0,1) = (1,1,1)

2. ## Re: Dimensions of linear subspaces

(i) fairly obviously has dimension 1, not 2. Was that a typo? As for (ii), in what sense is {(1, 2, 3, 4, ...)} a function from N to R?

For (iii), Are the two vectors independent? That is, is one a multiple of the other? Certainly $\displaystyle 2*\overline{1}= \overline{2}$. Is $\displaystyle 2*\overline{2}= \overline{1}$?

3. ## Re: Dimensions of linear subspaces

i) oh yea sorry it was a typo indeed.

ii) Well the span is a subset of the function that goes from N to R. Dont know more than this

iii) Well 2*1 = 2 and 2*2 = 1. But I cant tell why you do 2*2 in the first place and how to get a conclusion from this?
Additionally 1+2 = 0 and 2+1 = 0. This would mean its linear dependant, right? But does this help? Im just throwing out my thoughts here.

4. ## Re: Dimensions of linear subspaces

Originally Posted by Cyganek
i) oh yea sorry it was a typo indeed.

ii) Well the span is a subset of the function that goes from N to R. Dont know more than this
That makes no sense. The "span" is a vector space, not a function.

iii) Well 2*1 = 2 and 2*2 = 1. But I cant tell why you do 2*2 in the first place and how to get a conclusion from this?
Additionally 1+2 = 0 and 2+1 = 0. This would mean its linear dependant, right? But does this help? Im just throwing out my thoughts here.
That is the whole point. Yes, $2(\overline{1}, \overline{0}, \overline{2})= (\overline{2}, \overline{0}, \overtime{1})$ so one is a multiple of the other- they are dependent. That is equivalent to just one of the two vectors so this span is one dimensional.