Finite collection of linear functionals can be separated

32 Views Asked by At

Suppose $V$ is a finite-dimensional $k$-vector space, consider a finite collection $\{\alpha_1, \dots, \alpha_n\}\subset V^*$ of linear functionals on $V$. Then there exists a vector $v\in V$ such that $\alpha_i(v)\ne \alpha_j(v)$ whenever $i \neq j$.

Is this true? I believe this is true but not sure how to prove it. If not is there a counter example?