Respuesta :
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 : [tex](x_1,x_2)+(y_1,y_2)=(0,x_2+y_2)[/tex]
Definition of multiplication : [tex]\alpha(x_1,x_2)=(\alpha x_1,\alpha x_2)[/tex]
We need to check which axioms of the vector space satisfied by V.
Let, [tex](x_1,x_2)=u\:\:,\:\:(y_1,y_2)=v\:\:,\:\:(z_1,z_2)=w\:\epsilon R^2[/tex]
and α , β are scalars.
Axiom 1). [tex]u+v=(x_1,x_2)+(y_1,y_2)=(0,x_2+y_2)\epsilon R^2[/tex]
Axiom 2). [tex]u+v=(x_1,x_2)+(y_1,y_2)=(0,x_2+y_2)[/tex]
[tex]v+u=(y_1,y_2)+(x_1,x_2)=(0,y_2+x_2)=(0,x_2+y_2)[/tex]
⇒ u + v = v + u
Axiom 3). [tex](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)[/tex]
[tex]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)[/tex]
⇒ ( u + v ) + w = u + ( v + w )
Axiom 4). [tex]u+0=(x_1,x_2)+(0,0)=(0,x_2+0)=(0,x_2)[/tex]
⇒ u + 0 ≠ u
Axiom 5). [tex]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)[/tex]
⇒ u + (-u) = 0
Axiom 6). [tex]\alpha(u)=\alpha(x_1,x_2)=(\alpha x_1,\alpha x_2)\epsilon R^2[/tex]
Axiom 7). [tex]\alpha(u+v)=\alpha(0,x_2+y_2)=(0,\alpha(x_2+y_2))=(0,\alpha x_2+\alpha y_2)[/tex]
[tex]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)[/tex]
⇒ α ( u + v ) = αu + αv
Axiom 8). [tex](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)[/tex]
[tex]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)[/tex]
⇒ ( α + β )u ≠ αu + βu
Axiom 9). [tex](\alpha.\beta)u=\alpha\beta(x_1,x_2)=\alpha(\beta x_1,\beta x_2)=\alpha(\beta u)[/tex]
⇒ ( αβ )u = α( βu )
Axiom 10). [tex]1.u=1(x_1,x_2)=(x_1,x_2)[/tex]
⇒ 1 . u = u
Therefore, Given Set is not a vector space as Axiom 4 and 8 are not satisfied.
