# 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?

• If $T^i = 0$, then the vectors $T^i v,\, T^{i+1}v,\,\dotsc,\, T^n v$ are all $0$, then the system is surely linearly dependent too. – Daniel Fischer Mar 2 '14 at 22:11
• If those are all zero, then the set of $T^jv$ for $0\leq j<i$ might be linearly independent, right? Then adding in those zero vectors shouldn't make it linearly dependent. – ShadesOfGrey Mar 2 '14 at 22:33
• Any set of vectors containing a zero is linearly dependent. You then have the nontrivial relation $\vec{0} = 0\cdot v_1 + \dotsc + 0\cdot v_j + 1\cdot \vec{0} + 0\cdot v_{j+2} + \dotsc + 0\cdot v_k$. – Daniel Fischer Mar 2 '14 at 22:37

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.
• But, according to the author's proof, does that mean every Vector $v$ in $V$ is then an Eigen Vector? – MathMan Jan 26 at 9:54