Vector space questions?

1. well addition is closed. the trick here is to see if we can find a suitable identity element and inverses for this non-standard addition.

so we need to see if we can find some element (a,b,c) in R^3 with:

(x1+a+1,y1+b+1,z1+c+1) = (x1,y1,z1) for all (x1,y1,z1) in R^3.

it appears that (-1,-1,-1) will do the trick.

now we need to see if we can find an inverse for (x1,y1,z1), that is,

some vector (u,v,w) with (x1,y1,z1) + (u,v,w) = (-1,-1,-1).

now (x1,y1,z1) + (u,v,w) = (x1+u+1,y1+v+1,z1+w+1), so we need:

x1+u+1 = -1, or u = -x1 - 2

y1+v+1 = -1, so v = -y1 - 2

z1+w+1 = -1, so w = -z1 - 2.

thus the additive inverse of (x1,y1,z1) is (-x1-2,-y1-2,-z1-2).

now we need to show that this new addition is associative and commutative:

(x1,y1,z1) + ((x2,y2,z2) + (x3,y3,z3)) = (x1,y1,z1) + (x2+x3+1,y2+y3+1,z2+z3+1)

= (x1+(x2+x3+1)+1,y1+(y2+y3+1)+1,z1+(z2+z3+1)+1)

= ((x1+x2+1)+x3+1,(y1+y2+1)+y3+1,(z1+z2+1)+z3+1)

= (x1+x2+1,y1+y2+1,z1+z2+1) + (x3,y3,z3)

= ((x1,y1,z1) + (x2,y2,z2)) + (x3,y3,z3)

for commutativity: (x,y1,z1) + (x2,y2,z2) =

(x1+x2+1,y1+y2+1,z1+z2+1) = (x2+x1+1,y2+y1+1,z2+z1+1)

= (x2,y2,z2) + (x1,y1,z1)

so R^3 with this new "+" is an abelian group, but we're not home-free yet. we need to show that

(c1+c2)(x,y,z) = c1(x,y,z) + c2(x,y,z).

(c1+c2)(x,y,z) = ((c1+c2)x,(c1+c2)y,(c1+c2)z)

= (c1x+c2x,c1y+c2y,c1z+c2z), but:

c1(x,y,z) + c2(x,y,z) = (c1x,c1y,c2z) + (c2x,c2y,c2z) =

(c1x+c2x+1,c1y+c2y+1,c1z+c2z+1), which is not the same

(if you have doubts, let (x,y,z) = (1,1,1), c1 = c2 = 1, then

(1+1)(1,1,1) = 2(1,1,1) = (2,2,2), while (1)(1,1,1) + (1)(1,1,1) =

(1,1,1) + (1,1,1) = (3,3,3), they are not the same).

so this is not a vector space, one of the axioms fails.

2. we have already seen that the addition works, but here we have a different multiplication, which may work. let's pick up with the axiom that failed before:

(c1 + c2)(x,y,z) = ((c1+c2)x + (c1+c2) - 1,(c1+c2)y + (c1+c2) - 1, (c1+c2)z + (c1+c2) - 1)

= (c1x+c2x+c1+c2-1,c1y+c2y+c1+c2-1,c1z+c2z+c1+c2-1)

= (c1x+c1-1+c2x+c2-1+1,c1y+c1-1+c2y+c2-1+1,c1z+c1+c2z+c2-1+1)

= (c1x+c1-1,c1y+c1-1,c1z+c1-1) + (c2x+c2-1,c2y+c2-1,c2z+c2-1)

= c1(x,y,z) + c2(x,y,z). groovy.

next, we need to show c((x1,y1,z1) + (x2,y2,z2)) = c(x1,y1,z1) + c(x2,y2,z2).

c((x1,y1,z1) + (x2,y2,z2)) = c(x1+x2+1,y1+y2+1,z1+z2+1)

= (c(x1+x2+1)+c-1,c(y1+y2+1)+c-1,c(z1+z2+1)+c-1)

= (cx1+cx2+2c-1,cy1+cy2+2c-1,cz1+cz2+2c-1)

= (cx1+c-1+cx2+c-1+1,cy1+c-1+cy2+c-1+1,cz1+c-1+cz2+c-1+1)

= (cx1+c-1,cy1+c-1,cz1+c-1) + (cx2+c-1,cy2+c-1,cz2+c-1)

= c(x1,y1,z1) + c(x2,y2,z2), done that.

the next axiom is going to get messy: we need to show that:

c1(c2(x,y,z)) = (c1c2)(x,y,z)

c1(c2(x,y,z)) = c1(c2x+c2-1,c2y+c2-1,c2z+c2-1)

= (c1(c2x+c2-1)+c1-1,c1(c2y+c2-1)+c1-1,c1(c2z+c2-1)+c1-1)

= (c1c2x+c1c2-c1+c1-1,c1c2y+c1c2-c1+c1-1,c1c2z+c1c2-c1+c1-1)

= ((c1c2)x+c1c2-1,(c1c2)y+c1c2-1,(c1c2)z+c1c2-1)

= (c1c2)(x,y,z) whew!

the rest of the axioms are easy to show:

1(x,y,z) = (1x+1-1,1y+1-1,1z+1-1) = (x,y,z)

0(x,y,z) = (0x+0-1,0y+0-1,0z+0-1) = (-1,-1,-1) (which although it looks odd,

is the "0-vector" of our funky space).

so having shown that (2) satisfies all the axioms for a vector space, we can conclude that R^3 with this "nonstandard" vector addition and scalar multiplication does indeed form a vector space.