Does the Set T Form a Vector Space Under Given Operations?

[hahatyshka]

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..??

 

[Office_Shredder]

xyz=0 means when you multiply the three numbers together you get 0. In particular this tells you one of x,y or z must be zero (think about why if you can't see it). Is this closed under addition?

 

[HallsofIvy]

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')

You haven't done anything but state what you want to prove! In particular, you haven't shown that if (x,y,z) and (x',y',z) are in V then (x+x', y+ y', z+ z') is in V. If xyz= 0 and x'y'z'= 0, does it follow that (x+x')(y+y')(z+z')= 0?
(Hint, just use "0"s and "1"s for x, y, z, x', y', 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..??

 

[hahatyshka]

so would this solution be correct?

so if xyz=0 then x or y or z or (x,y,and z) must equal zero
suppose z=0
axiom 10: 1(x,y,z)=(1x,1y,1z)=(1x,1y,0)=(x,y,0)
which doesn't equal (x,y,z)
therefore not a vector space

please help..!

 

[Office_Shredder]

Well, it does because z=0.

Try taking two sample vectors and adding them together

 

[John D.]

This is an old thread but I just ran into the same question myself, it's problem 1.3(c) from Hassani.
Basically a space with the condition that xyz=0 includes all vectors in the three planes: xy, yz and xz, because any product of xy, yz or xz is allowed because the third coordinate is zero so xyz=0 is satisfied. Now, if you take a vector from the xy plane and add it to a vector in the xz or the yz plane you're obviously NOT going to get something that is still in one of those three planes. Unless of course those two vectors lie on two of the axes, but in general the answer is no.