What Defines the Zero Vector in Modified Vector Space Operations?

[v1ru5]

Yes in $R^2$ but not in your problem. That isn't your rule for scalar multiplication.

Oh okay so

k~(x, y) = (kx - k + 1, ky + 2k -2)
-1~(x,y) = -x + 1 + 1, -y -2 - 2
= -x + 2, -y - 4

So -(x, y) = (-x + 2, -y - 4)

and that is correct since I got that as my answer when I did it the other way. So I can conclude that if I want to get the additive inverse of any vector, I can multiply that vector by scalar $-1$ using the rules for scalar multiplication given.

 

[LCKurtz]

Yes that is correct. I don't think we need to beat this horse any more. But using these techniques you could verify all the axioms of a vector space work even though its arithmetic rules may seem bizarre. To paraphrase, "This ain't your daddy's vector space".

 

[v1ru5]

Yes that is correct. I don't think we need to beat this horse any more. But using these techniques you could verify all the axioms of a vector space work even though its arithmetic rules may seem bizarre. To paraphrase, "This ain't your daddy's vector space".

Thanks for your all your help Professor LCKurtz.

 

[s_nirmit]

how do you find the -v vector then ?

 

[s_nirmit]

what is the correct awnser for -v

 

[LCKurtz]

The additive inverse of (x,y) is the vector (a,b) you can ⊕ to (x,y) and get the additive identity.

how do you find the -v vector then ?

The answer is in the thread or you can do it yourself starting with the above statement.

 

[s_nirmit]

is it -v= (1-x), (-2-y) ?

 

[LCKurtz]

is it -v= (1-x), (-2-y) ?

No. Read the thread and/or show your work.