Question #68955

Let V=R2 . Define addition + on V by ( x1 , y1 ) + ( x2 , y2 ) = ( x1+x2 , y1+y2 ) and scalar multiplication . by r . (a , b) = (ra , 0). Check whether V satisfies all the
conditions for it to be a vector space over R with respect to these operations.
1

Expert's answer

2017-06-21T15:03:06-0400

Answer on Question #68955 – Math – Linear Algebra

Question

Let V=R2V = \mathbb{R}^2. Define addition + on VV by


(x1,y1)+(x2,y2)=(x1+x2,y1+y2)(x_1, y_1) + (x_2, y_2) = (x_1 + x_2, y_1 + y_2)


and scalar multiplication \cdot by


r(a,b)=(ra,0).r \cdot (a, b) = (ra, 0).


Check whether VV satisfies all the conditions for it to be a vector space over R\mathbb{R} with respect to these operations.

Solution

VV is not a vector space over R\mathbb{R} with respect to these operations. The axiom of identity element of scalar multiplication is not fulfilled.

In order to be a vector space, VV has to fulfill the axiom:


1(a,b)=(a,b)1 \cdot (a, b) = (a, b)


for arbitrary (a,b)V(a, b) \in V, where 1 denotes the multiplicative identity in R\mathbb{R}.

But for example, for (a,b)V(a, b) \in V with b0b \neq 0 for arbitrary rRr \in \mathbb{R} we have


r(a,b)=(ra,0)(a,b).r \cdot (a, b) = (ra, 0) \neq (a, b).


Therefore, VV is not a vector space over R\mathbb{R} with respect to these operations.

**Answer**: VV is not a vector space over R\mathbb{R} with respect to these operations.

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