Question

Proving that the following is not a vector space:
First octant of (x,y,z) space. I figured that it's not a vector space, but I can't figure out what properties other than the additive inverse property that it fails. Everything else seems to be right.

Answer 1

It's sufficient that any property fail. Call the octant E. The fact that v in E does not imply v in E is sufficient for E not to be a subspace.

For the record, scalar multiplication also fails if the scalar is negative.

Answer 2

THANKYOUU. ;_____; I kept going in circles through the list of properties and couldn't figure out what else it failed. It's a multi choice answer and I'm supposed to pick all the properties that it fails. I didn't understand what you meant by "The fact that v in E does not imply v in E is sufficient for E not to be a subspace." though.

I'm not familiar enough with vectors and vector spaces to pick out good counter cases x-x;

Answer 3

Take (1,1,1). Then the vector (-1,-1,-1) is not in E

Answer 4

By the way, is the definition of E that all the ordinates are positive? If so, then the zero vector isn't in E either.

Answer 5

Nope, it wasn't defined to be positive. My previous question asking about if they were all positive was just to see if I could figure it out on my own :c cause I kept second guessing and wondering if there was any way to add two vectors together to get something that was outside the octant.

E is 0 inclusive, so it did have a zero vector.

But yea! Understood.