Sheldon Axler's proof that every operator on a complex vector space has an eigenvalue

Question

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 !

Answer 1

We can have, for example, $(T-\lambda_mI)v=u\neq0$ and $(T-\lambda_{m-1}I)u=0$.

Answer 2

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 $T-\lambda_i I$.

Answer 3

The last statement does not mean that you can find such $i$ that $(\hat{T}-\lambda_i \hat{I})\vec{v} = 0$ for any $\vec{v}$. It rather means, that you can decompose you vector $$ \vec{v} = a_1 \vec{u}_1 + \dots + a_m \vec{u}_m $$ so that every $u_i$ can find "its own" $(\hat{T}-\lambda_i \hat{I})$ and become zero.