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

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 !

• 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. – user1551 Jan 26 at 16:35
• The proof doesn't show that $v$ is an eigenvector... – David C. Ullrich Jan 26 at 19:43

## 3 Answers

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

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 $$T-\lambda_i I$$.

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.