Question regarding Sheldon Axler's proof that every operator on a complex vector space has an eigenvalue For reference, Axler's proof was copied in full in this question.  In the beginning of that proof, Axler chooses the nonzero vector $v$ without loss of generality.  This made me doubt his proof because it is obvious that not all nonzero vectors are necessarily eigenvectors.  So I looked for something that would help me to distinguish $v$ from other arbitrary vectors, and I realized that for it to be true that $(v,Tv,\dots,T^nv)$ is linearly dependent, each $T^i$ for $0<i\leq n$ must be nonzero.  Thus I am left thinking that Axler's proof must show that such a $v$ can definitely be chosen.  Was this step left out of the proof because the existence of such a $v$ trivial in some manner that I am not noticing?  Or is my reasoning about the proof incorrect?
 A: Well, the proof doesn't claim that $v$ is an eigenvector, so there's no problem if it isn't.
The vectors $(v,Tv,\dots,T^nv)$ are linearly dependent regardless of the values of $v$ and $T$. It doesn't matter whether $v$ is nonzero. It doesn't matter whether $T^i$ is nonzero. It doesn't even matter that $T$ is a linear operator. Any set of $n+1$ vectors is linearly dependent.
A: I believe the point of confusion is the assumption
$$0=c(T-\lambda_1 I)\cdots (T- \lambda_m I)v$$
means $(T-\lambda_k I)v = 0$ for some $v$. Given this assumption, it would indeed seem to be the case that $v$ must be an eigenvector for the eigenvalue $\lambda_k$.
However, recall the way multiplication of operators is defined here: The multiplication of the operators $T * T = T^2$ means $(TT)v = T(Tv)$, not $(TT)v = (Tv)(Tv)$, as we can't multiply two vectors in a vector space, only a vector with a scalar. Function multiplication is defined as function composition.
That being said, the main equation is saying we plug $v$ into the operator $T-\lambda_m I$, then take the output and plug it into $T-\lambda_{m-1}I$, etc. until we finally plug the output into $T - \lambda_1 I$ and get $0$. Since we start with a nonzero input and get a zero output, there is some $\lambda_k$ such that $T - \lambda_k I$ takes in a nonzero input and produces a zero output, hence $T - \lambda_k I$ is not invertible and $\lambda_k$ is an eigenvalue.
We cannot definitively conclude anything about the starting vector $v$, since it may "disappear" after the first plugin into $T- \lambda_m I$. For example, say $(T-\lambda_m I)v$ is equal to an eigenvector for the eigenvalue $\lambda_{m-1}$, then when we plug in this eigenvector into the next operator $T - \lambda_{m-1} I$, we get $0$. Note that we can't conclude $v$ is an eigenvector.
