Since the proof is valid for any $v \in V$, the proof makes sure that any vector $v$ is an eigenvector which is of course not true. What is the error in this line of thought? Thanks a lot !
Sheldon Axler’s proof that every operator on a complex vector space has an eigenvalue
eigenvalues-eigenvectorslinear algebra
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Best Answer
At the end of the proof it is only asserted that $T-\lambda_i I$ is not injective for some $i$. It does not give you $(T-\lambda_i I)v=0$ and we cannot say that $v$ is eigen vector corresponding to $T-\lambda_i I$.