Define in F R2017 as the set of all vectors x1 x2 x2017 tha

Define in F R^2017 as the set of all vectors (x_1, x_2, ..., x_2017) that satisfy 1x_1 + 2x_2 + 3x_3 + middot middot middot + 2017_x_2017 = 0. Show F is a subspace of dimension 2016 in R^2017.

Solution

Define T : R^2017-----------> R

by T (x)=0 ,x in R^2017

Ker T ={ x in R^2017 : T(x)=0} =

{x in R^2017:x1+2x2+3x3+....2017x2017=0} which is the given subspace

by rank nulli theorm

RankRankT+dmm kerT=dim R^2017=2017

and since R(T) is suspace of R then rankT=1

dim kerT=2017-1=2016

hencec the dim of the given suspac is 2016