- #1
hahatyshka
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Homework Statement
Let T be the set of all ordered triples of real numbers (x,y,z) such that xyz=0 with the usual operations of addition and scalar multiplication for R^3, namely,
vector addition:(x,y,z)+(x',y',z')=(x+x',y+y',z+z')
scalar multiplication: k(x,y,z)=(kx,ky,kz)
Determine whether T, under the operations of addition and scalar multiplication given above, forms a vector space.
Homework Equations
The Attempt at a Solution
I think that when there are usual addition and multiplication that this would be a vector space, but I am not sure how am I suppose to show it using the axioms. Also, I'm sure what xyz=0 really means?
However, would this solution be correct..?
Axiom 1: If u vector exists in V and v vector exists in V then u+v exist in V.
: (x,y,z)+(x',y',z')= (x+x',y+y',z+z')
Axiom 2: u+v=v+u
:(x,y,z)+(x',y',z')=(x',y',z')+(x,y,z)=(x'+x,y'+y,z'+z)=(x+x',y+y',z+z')
Axiom 3: u+(v+w)=(u+v)+w
:this would hold b/c usual addition
Axiom 4: There exists a 0 vector such that u+0=u=0=u for all vectors u in V.
:(x,y,z)+(0,0,0)=(x,y,z) -> so, (0,0,0)=0 vector
Axiom 5: For every u in V there exists a -u in V such that u+(-u)=0=(-u)+u.
:(x,y,z)+(-x,-y,-z)=(0,0,0) -> so, (-x,-y,-z) is the -u vector
Axiom 6: If k exists in R(real number) and u exist in V then ku exist in V.
:k(x,y,z)=(kx,ky,kz)
Axiom 7: k(u+v)=ku+kv , k is a real number
:k[(x,y,z)+(x',y',z')]=k(x,y,z)+k(x',y',z')
Axiom 8: (k+l)u=ku+lu , k and l are real numbers
: (k+l)(x,y,z)=k(x,y,z)+l(x,y,z)
Axiom 9: k(lu)=(kl)u , k and l are real numbers
: k[l(x,y,z)]=(kl)(x,y,z)
Axiom 10:1u =u
: 1(x,y,z)=(1x,1y,1z)=(x,y,z)
is this correct? but i don't get where am i suppose to use xyz=0..??