# Vector space

1. Oct 8, 2007

### skydiver_spike

rather than use the standard definitons of addition and scalar multiplication in R^2, suppose these two operations are defined as follows.
(x1,y1)+(x2,y2)=(x1,0)
c(x,y)=(cx,y)

what would the zero vector be? can it be (0,-y1)? why or why not? I think it should fail this axiom u+0=u.

2. Oct 8, 2007

### radou

First of all, your addition, as defined, is not commutative, i.e. (x1, y1) + (x2, y2) = (x1, 0) $\neq$ (x2, y2) + (x1, y1) = (x2, 0).

3. Oct 8, 2007

### HallsofIvy

Staff Emeritus
That is simply not a vector space. Not only is addition not commutative, as radou said, worse, there simply is NO additive identity. The equation (x1,y1)+ (x,y)= (x1,0)= (0,0) simply has no solution (x,y). This set is not a group with that addtion operation.

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