Hmm...starting to do Vector spaces and such, and i'm having a bit of trouble!
Why does:
$\displaystyle R^3: x + y + z = 0; x - y + 2z = 8 ?$
Not have a Zero vector?
Surely all vectors have Zero vectors as you can just add (0, 0, 0) ?
Thankyou
Hmm...starting to do Vector spaces and such, and i'm having a bit of trouble!
Why does:
$\displaystyle R^3: x + y + z = 0; x - y + 2z = 8 ?$
Not have a Zero vector?
Surely all vectors have Zero vectors as you can just add (0, 0, 0) ?
Thankyou
I believe you are asking about the subset of $\displaystyle R^3$, that is all vectors of the form <x, y, z>, such that x+ y+ z= 0 and such that x- y+ 2z= 8. That is, the set of all <x, y, z> that satisfy both equations. That is, in fact, the intersection of the sets of vectors that satisfy each equations separately.
That obviously does not include the 0 vector, <0, 0, 0>, because it does not satisfy the second equation: 0 - 0+ 2(0)= 0, not 8. You could also argue that it is not a subspace because it does not "closed under scalar multiplication". If <x, y, z> satisfies x- y+ 2z= 8 and n is any number other than 1, then <nx, ny, nz> satisfies nx- ny+ 2(nz)= n(x- y+ 2z)= 8n, not 8. One could also argue that it is not closed under addition. If <x, y, z> and <a, b, c> are two vectors not satisfying x- y+ 2z= 8 and a- b+ 2c= 8, then their sum, <x+ a, y+ b, z+ c> satisifies (x+ a)- (y+ b)+ 2(z+ c)= (x- y+ 2z)+ (a- b+ 2c)= 8+ 8= 16, not 8.