# Is this a vector space

• Dec 23rd 2012, 08:05 AM
scnakandala
Is this a vector space
A = {(x,y)| x,y belong R}
scalar multiplication and addition are defined as
a(x,y) = (ax,ay)
(x1 , y1) + (x2 , y2) = (x1 + x2 + 1, y1+y2+1)

1) can we define zero vector as (-1,-1)
2) can we define -(x,y) as (-x-1,-y-1)
3) is A a vector space
• Dec 23rd 2012, 09:45 AM
emakarov
Re: Is this a vector space
Good questions. Why don't you try answering them?
• Dec 23rd 2012, 09:53 AM
HallsofIvy
Re: Is this a vector space
Your scalar mutiplication must satisfy a(u+ v)= au+ av. Is that true?
• Dec 23rd 2012, 10:11 AM
scnakandala
Re: Is this a vector space
yes
• Dec 23rd 2012, 10:14 AM
scnakandala
Re: Is this a vector space
I don't see any reason why first and second statements are wrong.
But the theorems -1u = -u and 0u = 0 contradicts my answer.
• Dec 23rd 2012, 10:31 AM
emakarov
Re: Is this a vector space
Quote:

Originally Posted by HallsofIvy
Your scalar mutiplication must satisfy a(u+ v)= au+ av. Is that true?

Quote:

Originally Posted by scnakandala
yes

Quote:

Originally Posted by scnakandala
I don't see any reason why first and second statements are wrong.

Let's see. "Can we define -(x,y) as (-x-1,-y-1)?" If yes, then -(x, y) + (x, y) = (-x - 1, -y - 1) + (x, y) = (-x - 1 + x + 1, -y - 1 + y + 1) = (0, 0). Have you done this computation? How does it square with the answer to question 1?

The definition for unary minus can be adjusted so that -(x, y) + (x, y) = (-1, -1). However, post #3 points out a real problem.

Quote:

Originally Posted by scnakandala
But the theorems -1u = -u and 0u = 0 contradicts my answer.

Yes, -1(x, y) = (-x, -y) according by definition, but then -1(x, y) + 1(x, y) ≠ 0(x, y).