# Can the zero vector be (1,1)?

Given the operation in $$\mathbb{R}^2$$:

$$(x_1,y_1) + (x_2,y_2) = (x_1x_2,y_1y_2)$$

I would like to find whether this is a vector space in $$\mathbb{R}$$. Looking at the Additive Zero Axiom, we get:

$$(x_1,y_1) + \boldsymbol{0} = (x_1(0),y_1(0)) = \boldsymbol{0}$$

To satisfy the Additive Zero Axiom, $$(x_1,y_1) + \boldsymbol{0} = (x_1,y_1)$$ must be true. For this to be true, $$\boldsymbol{0}$$ would have to be $$(1,1)$$

Is this possible, or would we be able to say this is not a vector space?

• By far nothing's gone wrong. Go ahead and check the other axioms Sep 16 at 3:13
• Welcome to the site! If the issue has been resolved, accepting and/or upvoting answers is the best way to say "thanks!": it scores points, signals resolution, and prevents bumping and automatic deletion. 2 days ago

Is not a vector space, for instance let us try find the zero element in $$\mathbb{R}^2$$ with the given operation.

Let $$P=(x,y)\in \mathbb{R}^2$$

$$(x,y)+(e_1,e_2)=(x,y)$$ implies $$(xe_1,ye_2)=(x,y)$$ and then $$e_1=1$$ and $$e_2=1$$ it is the zero element must be $$(1,1)$$.

But in this case $$(0,0)$$ isn´t invertible since $$(0,0)+(a,b)=(1,1)$$ implies $$(0,0)=(1,1)$$ which is a contradiction.

• Thank you for your help! This now makes complete sense. Sep 16 at 2:27
• You´re welcome! Sep 16 at 2:28

The additive identity is indeed $$(1,1)$$.

Let's check for inverse of $$(0,0)$$.

For any $$x, y \in \mathbb{R}$$,

$$(0, 0) + (x, y)= (0,0) \ne (1,1).$$ Hence it can't be a vector space.

• Thank you very much for your reply! This makes sense now, (0,0) does not have an additive inverse with (1,1) being the additive identity Sep 16 at 2:26