# Linearly independent eigenvector

Question: Suppose $v_1,\cdots,v_m$ are linearly independent vectors in $V$. Prove that there exists $T\in\mathcal{L}(V)$ such that $v_1,\cdots,v_m$ are eigenvectors of $T$ corresponding to distinct eigenvalues.

My Proof: Since $(v_1,\cdots,v_m)$ are linearly independent, it can be extended to a basis $(v_1,\cdots,v_m,w_1,\cdots,w_n).$ Define an invertible linear map $T$ by $$T(a_1v_1+\cdots+a_mv_m+b_1w_1+\cdots+b_nw_n)=a_1\lambda_1v_1+\cdots+a_m\lambda_mv_m$$for $\lambda_i\neq \lambda_j$ with $1\leq i\leq j\leq m.$ Then this is the linear map such that $v_1,\cdots,v_m$ are eigenvectors of $T$ corresponding to distinct eigenvalues.

Is this correct?

If you are working over an infinite field then, yes, it is correct. Otherwise it isn't because, in some cases, you may be unable of finding $m$ distinct $\lambda_k$'s.