Question

consider: R^3 is a vector space, V =(x, y, z) is a subset of R^3
if x =0, is V is a vector space?
Please, check

Answer 1

my work is
let k is a scalar
u =(x,y, z)= (0, y, z)
ku = (0, ky, kz ) !=(k.0, ky, kz) so it is not closed under multiplication axiom
--> V is not VS?

Answer 2

how is k.0 != 0 ?

Answer 3

since V =(x, y, z)
so any vector in V should be closed under multiplication to get the form ku = (kx,ky,kz) , but 0 ??

Answer 4

so is T(ku) = kT(u)?

Answer 5

@sourwing is it not we just check + and* axioms?

Answer 6

yes, we have to check both

Answer 7

@sourwing what is the role of T here?

Answer 8

let x=0, then <kx,ky,kz>=<k(0),ky,kz>=<0,ky,kz>=k<0,y,z>

Answer 9

just the name for the transformation

Answer 10

so I say it is closed under scalar multiplication, though I never did have a proper linear algebra course

Answer 11

yes, it's closed under multiplication. What about addition?
Is T(u+v) = T(u) + T(v)

Answer 12

Yes, to algebra course, it 's quite easy to prove.But this is Abstract algebra, so. I confused everything

Answer 13

@sourwing yes it is , as TuringTest say, I believe it

Answer 14

one more question: if R ^3 is C^3 ,then, it turns true?

Answer 15

since x =0 and 0 is considered as a complex number , too. right? so , it is a VS, too, right?
and what if x is real on C^3 with V = (real, complex, complex)? is V a Vector space?

Answer 16

that I dont know :D

Answer 17

yeah that's where it gets pretty technical...
all real numbers are a subset of the complex numbers

Answer 18

and a subset can not be a subspace if it doesn't have 0 vector there, right?

Answer 19

yes

Answer 20

V = (real, complex, complex) should be closed under scalar multiplication, but under addition the complex components could have their imaginary part=0, making them "real", but again, all reals are a subset of C, so...

Answer 21

Thanks a lot. This is my first exercise and I have to submit it to collect my credit. I don't want to lose my credit because of those easiest stuff of the course.

Answer 22

I'm actually taking my first "real" linear algebra course now, with some abstract in it, so we'll be on this journey together it seems :)

Answer 23

yep, you should, it 's not hard, just go around and easy have mistake.

Answer 24

yeah I read a book on it. it's an interesting subject, but with things like method of cofactors and finding inverses of big matrices there is so much room for arithmetic or basic algebra errors. It can really get me frustrated. Hopefully I can focus more on the abstract part, seems less convoluted to me. Could be wrong about that though, hehe

Answer 25

:)