Prove that if a 1dimensional subspace W of Rn contains a non

Prove that if a 1-dimensional subspace W of Rn contains a nonzero vector with all nonnegative entries, then W contains a unique probability vector.

Solution

Let W be a 1-dimensional subspace of Rⁿ , and let p = ( 0,0,…,a_p,0,0,..,a_q,..,0) be the non-zero vector in W with non negative entries.. Then, the only other vector in W is the zero vector ( 0,0,.., 0,…0). Now, W is closed under scalar multiplication so that if is an arbitrary scalar, then p W. Further, p = ( 0,0,…,a_p,0,0,...,a_q,.,0) = ( 0,0,0,…, a_p,0,0,_,.. a_q..,0) . Since W contains only one non- zero vector p = ( 0,0,0,…,a_p,0,0,_,..,a_q..,0) , therefore, we must have a_p = a_p. and a_q = a_q Now, since is an arbitrary scalar, this is possible only if a_p = a_q = 0 for all a_p and a_{q .} However, this is a contradiction as p is a non-zero vector. The only solution, therefore is that only one a_p in p is non – zero and = 1 and a_p = 1. This means that the non-negative entries of p add up to one. And therefore, p is a probability vector.