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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 !

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    $\begingroup$ You can get rid of the vector $v$ if you don't like it. Since $\mathcal L(V)$ is $n^2$-dimensional, $I,T,T^2,\ldots,T^{n^2}$ cannot be linearly independent and $p(T)=0$ for some nonzero polynomial $p$ of degree $m\le n^2$. This does not make $m\le n$ as in Axler's proof, but the rest of the argument remains essentially the same. $\endgroup$ – user1551 Jan 26 at 16:35
  • $\begingroup$ The proof doesn't show that $v$ is an eigenvector... $\endgroup$ – David C. Ullrich Jan 26 at 19:43
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We can have, for example, $(T-\lambda_mI)v=u\neq0$ and $(T-\lambda_{m-1}I)u=0$.

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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 $T-\lambda_i I$.

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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.

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