Recent content by MoreDrinks

Thank you, that clears up a lot.

True, thanks! Would the correct term be that we're working in the "field" of R^3 or just R^3 space when talking about this?

If we're dealing with complex space, can scalars be complex? Thanks for the help!

Homework Statement Which of the following subsets of R3 are subspaces? The set of all vectors of the form (a,b,c) where a, b, and c are... Homework Equations 1. integers 2. rational numbers The Attempt at a Solution I think neither are subspaces. IIRC, the scalar just needs to be...

Thanks for all the help. I usually end up coming here when I'm exhausted and frustrated, and then we see sloppy mistakes on my part. You're a tremendous help.

Whoops, yeah, two. Thank you for your help. Oh, am I just wrong on the order of operations - the exponential would come first in kx^2, then multiplication by the scalar, which could leave us with negatives?

You are correct sir! Thanks.

Oh, okay. So, for example F(x)= x+1 G(x)= -x +1 Add them together and get a constant, 1. Is that correct? If you have a moment, what was wrong with my original reasoning with F(x)=x^2

Homework Statement Let V be the set of all nonconstant functions with operations of pointwise addition and scalar multiplication, having the real numbers as their domain. Is V a vectorspace? Homework Equations None. The Attempt at a Solution My guess is, no. For example F(x) =...

Thank you. Cheers.

Well, we start in R2 with the form (a,b2). Anything of the form (a,b2) that we can add to our original (a,b2) is going to leave us with a (a,b2). Any b in R2 that is squared is going to be positive, which doesn't leave us with a second element b2 that is negative and, subsequently, possessing...

Surely, Mark. Follow-up question: when you're checking to see if a subset of vectors is closed under scalar multiplication, must a scalar be within the original vector space from which the subset is derived?

It may take me a little longer than the other kids, but I get there. Thank you, sir.

Like, -1(a,b2)? So we get (-a,-b2)? Is this a problem - am I viewing the order of operations wrong or something?

As far as I can tell, yes, it's closed under addition and scalar multiplication, but the answer is that it's not a subspace. This is why I am stumped.