Answer to Question #205021 in Linear Algebra for Rohan Kumar

Question #205021

let V={(a,b,c)€R³|a+b=c} and W={(a,b,c)€R³|a=b} be subspaces of R³.Is R³ direct sum of V and W?


1
Expert's answer
2021-06-10T05:30:24-0400

Let V={(a,b,c)R3  a+b=c}V=\{(a,b,c)\in\R^3\ |\ a+b=c\} and W={(a,b,c)R3  a=b}W=\{(a,b,c)\in\R^3\ |\ a=b\} be subspaces of R3\R^3.


Since V={(a,b,c)R3  a+b=c}={(a,b,a+b)R3  a,bR}={a(1,0,1)+b(0,1,1)R3  a,bR},V=\{(a,b,c)\in\R^3\ |\ a+b=c\}=\{(a,b,a+b)\in\R^3\ |\ a,b\in\R\}=\{a(1,0,1)+b(0,1,1)\in\R^3\ |\ a,b\in\R\},

we conclude that the vector subspace VV is of dimension 2 with basis consisting of (1,0,1)(1,0,1) and (0,1,1).(0,1,1).


Taking into account that W={(a,b,c)R3  a=b}={(a,a,c)R3  a,cR}={a(1,1,0)+c(0,0,1)R3  a,cR},W=\{(a,b,c)\in\R^3\ |\ a=b\}=\{(a,a,c)\in\R^3\ |\ a,c\in\R\}=\{a(1,1,0)+c(0,0,1)\in\R^3\ |\ a,c\in\R\},

we conclude that the vector subspace WW is of dimension 2 with basis consisting of (1,1,0)(1,1,0) and (0,0,1).(0,0,1).


Since dimV+dimW=2+2=4>3=dimR3,\dim V+\dim W=2+2=4>3=\dim \R^3, we conclude that R3\R^3 can not be a direct sum of VV and W.W.



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