Answer to Question #205693 in Linear Algebra for Muzamel

Question

Show that R 3 is a real vector space. Show that the set {(x, y, 0)|x, y ∈ R} is subspace of R 3 .

Answer 1

$(1)\space proof:-\\ we\space will\space be\space proof\space that\space \space R_3\space is\space a\space real\space vector\space space.\\ definition\space of\space vector\space space\space \\ let\space F\space be\space a\space field\space and\space V\space be\space a\space set,\\ then\space \space V(F)\space called\space a\space vetor\space space\\ if\\ (V_1)\space (V,+)\space is\space commutative\space group\\ let\space \space a,b\space \isin\space F\space and\space \alpha,\beta\space \isin\space V\\ (V_2)\space a.(\alpha+\beta)=a.\alpha+a.\beta\\\space (V_3)\space (a+b).\alpha=a.\alpha+b.\alpha\\ (V_4)(ab).\alpha=a(b.\alpha)\\ (V_5)1.\alpha=\alpha\\ -------------------------------- according\space to\space question\space \\ V=R_3\space and\space F=R\space (given\space real\space vector\space space)\\ R_3\space is\space define\space as\space \\ R_3=\{\space (a,b,c)|a,b,c\space \isin\space R\space \}\\ now\\ (V_1)\space (R_3,+)\space is\space commutative\space group\\ (i)associate\space :-\\ let\space \alpha\space =(a_1,b_1,c_1),\space \beta\space =(a_2,b_2,c_2),\space \gamma\space =(a_3,b_3,c_3)\space \isin\space R_3\\ then\space easily\space we\space proof\space that\\ [(a_1,b_1,c_1)+(a_2,b_2,c_2)]+(a_3,b_3,c_3)=(a_1,b_1,c_1)+[(a_2,b_2,c_2)+(a_3,b_3,c_3)]\\ [\alpha+\beta]+\space \gamma\space =\alpha+[\beta+\space \gamma\space ]\\ (V_2)\space a.(\alpha+\beta)=a.\alpha+a.\beta\\\space a.[(a_1,b_1,c_1)+(a_2,b_2,c_2)]=a.[(a_1+a_2,b_1+b_2,c_1+c_2)]=[(a.a_1+a.a_2,a.b_1+a.b_2,a.c_1+a.c_2)]\\ a.[(a_1,b_1,c_1)+(a_2,b_2,c_2)]=[(a.a_1,a.b_1,a.c_1)+(a.a_2,a.b_2,a.c_2)]\\ a.[(a_1,b_1,c_1)+(a_2,b_2,c_2)]=a.(a_1,b_1,c_1)+a.(a_2,b_2,c_2)\\ a.(\alpha+\beta)=a.\alpha+a.\beta\\\space (V_3)\space (a+b).\alpha=a.\alpha+b.\alpha\\ (a+b).(a_1,b_1,c_1)=[(a+b).a_1,(a+b).b_1,(a+b).c_1]\\ (a+b).(a_1,b_1,c_1)=[(a.a_1+b.a_1),(a.b_1+b.b_1),(a.c_1+b.c_1)]\\ \space (a+b).(a_1,b_1,c_1)=a.(a_1,b_1,c_1)=+b.(a_1,b_1,c_1)\\ (V_4)(ab).\alpha=a(b.\alpha)\\ we\space can\space easily\space proof\space that\\ (ab).(a_1,b_1,c_1)=a(b.(a_1,b_1,c_1))\\ (V_5)1.\alpha=\alpha\\ it\space also\space can\space easily\space proof.\\ hence\space \space R_3\space is\space a\space real\space vector\space space.\\ (2)\space proof:-\\ we\space will\space be\space prove\space that\space set\space W={(x,\space y,\space 0)|x,\space y\space ∈\space R}\space is\space subspace\space of\space R_3\space .\\ let\space a,b\space \isin\space R\space \\ \alpha\space =(x_1,y_2,0)\space and\space \beta=(x_1,y_2,0)\space \space \isin\space W\space \\ for\space subspace\space we\space will\space proof\space this\space condition\\ a.\alpha\space +b.\beta\space \isin\space W\\ now\space start\\ a.\alpha\space +b.\beta=a.(x_1,y_2,0)\space +b.(x_1,y_2,0)\space \\ a.\alpha\space +b.\beta=(a.x_1,a.y_2,0)\space +(b.x_1,b.y_2,0)\space \\ a.\alpha\space +b.\beta=(a.x_1+b.x_1,a.y_2+b.y_2,0)\space \\ a.\alpha\space +b.\beta=((a+b).x_1,(a+b).y_2,0)\space \\ where\space (a+b).x_1,(a+b).y_2\space \isin\space R\space \\ hence\space \\ ((a+b).x_1,(a+b).y_2,0)\isin\space W\space \\ a.\alpha\space +b.\beta\space \isin\space W\space \\ hence\space \\ set\space W={(x,\space y,\space 0)|x,\space y\space ∈\space R}\space is\space subspace\space of\space R_3\space .\\$