Let $V$ be a vector space of dimension n. Show that there exists $n+1$ vectors $u_1,u_2,...,u_n,u_{n+1}$ such that every vector in $V$ can be expressed as a linear combination of $u_1,u_2,...,u_n,u_{n+1}$ with non-negative coefficients.
Can anyone check my solution? I checked my solution to the answer which was completely different so I was wondering if I could do it this way.
Dimension $n$ implies Basis has $n$ vectors.
Let $s$ be any vector in $V$ and let $T$ be a basis of $n$ vectors, i.e $T= (u_1,u_2,...,u_n)$. Therefore $s$ can be written as $s=c_1u_1+c_2u_2+...+c_nu_n$
Since $u_1,u_2,...,u_n,u_{n+1}$ are linearly dependent, let $u_{n+1}=c_1u_1+c_2u_2+...+c_nu_n$. Therefore $s=u_{n+1}=0u_1+0u_2+...+0u_n+1u_{n+1}$.
Therefore I have expressed every vector in V in terms of linear combinatyion of non-negative coefficients of 0 and 1!