Proof that a subset of R 2 is not a vector spaceHere is a subset of R 2 : A = [ ( x 1 , x 2 ) ∈ R 2 : x 1 − x 2 2 = 0 ]I am trying to prove that it is not a vector subspace.It is not empty, the (0,0) vector worksIt is closed under scalar multiplication (sign of x 1 and x 2 are changed but it works)I was a bit struggling to prove that it does not work for all x 1 and x 2 (Is it possible ?)I have chosen to take an u=(1,−1) and v=(16,4). Both of them are in R 2 . When I do u+v, I got a vector that is (17,3). This vector is not in A since x 1 − x 2 2 ≠ 0My question : Is there something wrong in my reasoning / solution ?

Question

Proof that a subset of             R        2   is not a vector spaceHere is a subset of             R        2  :  A  =  [  (      x    1    ,      x    2    )  ∈            R        2    :      x    1    −      x    2    2    =  0  ]I am trying to prove that it is not a vector subspace.It is not empty, the (0,0) vector worksIt is closed under scalar multiplication (sign of       x    1   and       x    2   are changed but it works)I was a bit struggling to prove that it does not work for all       x    1   and       x    2   (Is it possible ?)I have chosen to take an u=(1,−1) and v=(16,4). Both of them are in             R        2  . When I do u+v, I got a vector that is (17,3). This vector is not in A since       x    1    −      x    2    2    ≠  0My question : Is there something wrong in my reasoning / solution ?

salumeqi · Accepted Answer

I don&#039;t feel anything is wrong with your reasoning.In general, for any   a  ∈      R    ,  (      a    2    ,  ±  a  )  ∈  ALet   a  ≠  0.     (      a    2    ,  a  )  +  (      a    2    ,  −  a  )  =  (      a    2    ,  0  )  ∉  A  .  [  Closure Property Fails  ]

Proof that a subset of RR^2 is not a vector space

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