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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?

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1 Answer 1

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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.

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