Why operators on complex vector spaces have eigenvalues There is one thing I can't grasp about the proof given in the Linear Algebra Done Right book by Sheldon Axler (attached below).
In the last part it says that $(T - \lambda_1I)...(T - \lambda_mI)v = 0$, hence $T - \lambda_jI$ is not injective for some $j$.
What I don't understand is why the following reasoning is not correct:

*

*the factors in the equation can be reordered.

*suppose $\lambda_j$ is the only eigenvalue $T$ has. Let's put it at the end: $(T - \lambda_1I)...(T - \lambda_mI)(T - \lambda_jI)v = 0$

*the only way for the expression above to be equal to $0$ is if $(T - \lambda_jI)v = 0$ (because $\lambda_j$ is the only eigenvalue, so the other $T - \lambda_iI$ are injective).

*hence $v$ is an eigenvector of $T$ corresponding to the eigenvalue $\lambda_j$. But $v$ was chosen arbitrarily, so it can't be true.

I know that my logic is flawed but I can't see where. Would appreciate it if someone pointed out to me where I'm wrong.

 A: *

*If $f, g$ are polynomials over $\Bbb C$, then $f(T)g(T) = g(T)f(T)$ by merely definition.

*You do not know $\lambda_j$ could be an eigenvalue beforehand. Then items 2 to 4 can not proceed.

*The correct reasoning is simple: if no such $j$ exists, then all $\mathcal T - \lambda _j \mathcal I$ are injective, so $v = 0$ which is not what we have chosen, contradiction.

*Whenever you know an eigenvalue $c$, all corresponding eigenvectors must come from $\operatorname{Ker}(\mathcal T - c\mathcal I)$. In this proof Axler fixed a vector [which has the potential to be an eigenvector as he proved] first, rather than found some value before proof.

UPDATE
As for the flaw, even if $\lambda_j$ is the only eigenvalue, in the factorization you cannot claim that $\lambda _k \neq \lambda _j$ iff $j \neq k$. Like the answer you accepted stated, the expression can be something like $(T - \lambda_j I)^n$, then $(T-\lambda_j)^{n-4}v =0$ is also a possible case [say if $n \geqslant 4$], yet you asserted that $(T-\lambda_j)v$ is $0$, which is not necessarily true. A counterexample is given in the comment section of the other answer.
A: If $\lambda_j$ is the only eigenvalue then $(T−\lambda_j)^n$ is the full expression and reordering doesn't change anything.
A: 
the factors in the equation can be reordered.

The factors in the characteristic polynomial can be re-ordered, but factors can't in general be, and it's not entirely trivial to prove that they can be in this case.

hence v is an eigenvector of T corresponding to the eigenvalue λj. But v was chosen arbitrarily, so it can't be true.

What you seem to be getting at is that your conclusion is that all vectors are eigenvectors. It certainly can be the case that all vectors are eigenvectors of an operator. All vectors are eigenvectors of an operator that's a scalar multiple of the identity, and in one-dimensional space, all operators are such.
But okay, let's take n>1 and T be an arbitrary operator. Your confusion seems to be that you made an assumption (λ is the only eigenvalue), and arrived at a false statement. The conclusion is that the assumption is false. I'm not clear on what you think the problem is. For vector fields over algebraically complete fields (and $\mathbb C$ is algebraically complete), the number of eigenvalues (including multiplicity) is equal to the dimension (in fact, this can be proven by extending the very argument that you presented). So when you said "suppose $λ_j$ is the only eigenvalue T has", you're supposing something that (for dimension greater than 1) can't be the case, unless $λ_j$ is a multiplicity-n eigenvalue. You assumed something false, and came to a false conclusion. What's the problem?
