Question

Let V be the set of vectors in R2 with the following definition of addition and scalar multiplication: Addition [x1 x2] [y1 y2] = [0 x2 + y2] Scalar Multiplication: alpha [x1 x2] = [ax1 ax2] Determine which of the Vector Space Axioms are satisfied.

Answer 1

Answer:

Given Set is not a vector space as Axiom 4 and 8 are not satisfied.

Step-by-step explanation:

Given:

V is Vector  in R².

Definition of addition : $(x_1,x_2)+(y_1,y_2)=(0,x_2+y_2)$

Definition of multiplication : $\alpha(x_1,x_2)=(\alpha x_1,\alpha x_2)$

We need to check which axioms of the vector space satisfied by V.

Let, $(x_1,x_2)=u\:\:,\:\:(y_1,y_2)=v\:\:,\:\:(z_1,z_2)=w\:\epsilon R^2$

and α , β are scalars.

Axiom 1). $u+v=(x_1,x_2)+(y_1,y_2)=(0,x_2+y_2)\epsilon R^2$

Axiom 2). $u+v=(x_1,x_2)+(y_1,y_2)=(0,x_2+y_2)$

                $v+u=(y_1,y_2)+(x_1,x_2)=(0,y_2+x_2)=(0,x_2+y_2)$

⇒ u + v = v + u

Axiom 3). $(u+v)+w=(\:(x_1,x_2)+(y_1,y_2)\:)+(z_1,z_2)=(0,x_2+y_2)+(z_1,z_2)=(0,x_2+y_2+z_2)$

$u+(v+w)=(x_1,x_2)+(\:(y_1,y_2)+(z_1,z_2)\:)=(x_1,x_2)+(0,y_2+z_2)=(0,x_2+y_2+z_2)$

⇒ ( u + v ) + w = u + ( v + w )

Axiom 4). $u+0=(x_1,x_2)+(0,0)=(0,x_2+0)=(0,x_2)$

⇒ u + 0 ≠ u

Axiom 5). $u+(-u)=(x_1,x_2)+(\:-(x_1,x_2)\:)=(x_1,x_2)+(-x_1,-x_2)=(0,x_2+(-x_2)\,)=(0,x_2-x_2)=(0,0)$

⇒ u + (-u) = 0

Axiom 6). $\alpha(u)=\alpha(x_1,x_2)=(\alpha x_1,\alpha x_2)\epsilon R^2$

Axiom 7). $\alpha(u+v)=\alpha(0,x_2+y_2)=(0,\alpha(x_2+y_2))=(0,\alpha x_2+\alpha y_2)$

$a\lpha u+\alpha v=\alpha(x_1,x_2)+\alpha(y_1,y_2)=(\alpha x_1,\alpha x_2)+(\alpha y_1,\alpha y_2)=(0,\alpha x_2+\alpha y_2)$

⇒ α ( u + v ) = αu + αv

Axiom 8). $(alpha+\beta)u=(\alpha+\beta)(x_1,x_2)=((\alpha+\beta)x_1,(\alpha+\beta)x_2)=(\alpha x_1+\beta x_1,\alpha x_2+\beta x_2)$

$alpha u+\beta u=\alpha(x_1,x_2)+\beta(x_1,x_2)=(\alpha x_1,\alpha x_2)+(\beta x_1,\beta x_2)=(0,\alpha x_2+\beta x_2)$

⇒ ( α + β )u ≠ αu + βu

Axiom 9). $(\alpha.\beta)u=\alpha\beta(x_1,x_2)=\alpha(\beta x_1,\beta x_2)=\alpha(\beta u)$

⇒ ( αβ )u = α( βu )

Axiom 10). $1.u=1(x_1,x_2)=(x_1,x_2)$

⇒ 1 . u = u

Therefore, Given Set is not a vector space as Axiom 4 and 8 are not satisfied.