Find the one sentence that is wrong in the following proof.

Question

Here is the Proposition that is being proved:

Let $V$ be a finite dimensional vector space over $\mathbb C$ with dim $V = n \geq 1,$ and let $T \in \mathcal{L}(V).$ Then $T$ has an eigenvalue.

Here is the attempted proof:

Since dim $V = n \geq 1,$ there is a nonzero vector $v \in V.$ Consider the list $(v, Tv, T^2v, \dots , T^n v ),$ which has length $n + 1.$ Since dim $V = n,$ this list is linearly dependent. Thus there is a linear relation $$a_0v + a_1 Tv + a_2 T^2 v + \dots + a_n T^nv = 0$$ with not all $a_i$ being zero. Let $$p(z) = a_0 + a_1 z + \dots + a_n z^n \in \mathcal{P}(\mathbb C).$$

Then $p$ is not the zero polynomial and $p(T)v = 0.$ Let $m$ be the largest $i$ such that $a_i \neq 0;$ we have $p \in \mathcal{P}_m(\mathbb C).$ Since every polynomial $\mathcal{P}(\mathbb C)$ factors into linear factors, we may write $$p(z) = \alpha (z - \lambda_1)(z - \lambda_2) \dots (z - \lambda_m)$$ for some complex numbers $\alpha, \lambda_1, \lambda_2, \dots, \lambda_m.$ Thus $$p(T)v = (T - \lambda_1 I)(T - \lambda_2 I) \dots (T - \lambda_m I)v = 0,$$

where $I \in \mathcal{L}(V)$ is the identity operator.

If $(T - \lambda_i I)$ is injective for every i, then $p(T)$ is injective, since the composition of injective maps is injective.

But $p(T)v = 0,$ so $p(T)$ is not injective, so at least one $T - \lambda_i I$ is not injective. For this i, we have $(T - \lambda_i I)v = 0$ and so $v$ is an eigenvector of $T$ with eigenvalue $\lambda_i.$ Therefore, $T$ has an eigenvalue.

My questions are:

Is the statement I made bold above is the wrong statement in the proof? If so, why it is wrong? Is there a counterexample to it?

If the statement I made bold above is not the wrong statement in the proof, what is the wrong statement in the proof?

Could someone help me answer those questions please?

Answer 1

Wrong sentence:

For this $i$, we have $(T-\lambda_i I)v=0$.

This need not happen. The only guarantee as you claimed is one of the operators $T-\lambda_i I$ is not injective and hence its kernel is non-zero. That non-zero element in its kernel would be the eigen vector.

Answer 2

The mistake in your proof is that you don't know that your original $v$ is in the kernel of your non-invertible $T-\lambda_i I$. So, as @Yathi stated in their answer, the statement that is wrong is

For this $i$, $(T-\lambda_i I)v=0$

But that's not too bad of a mistake. You were correct up to that point, so "$(T-\lambda_i I)$ is not injective" is correct. So you can instead say "That means that there is some non-zero vector $w$ such that $(T-\lambda_i I)w=0$, so $\lambda_i$ is an eigenvalue of $T$".

(I'm assuming here that you're comfortable with injectivity and kernels of linear maps, and the different equivalent ways to show that something is an eigenvalue. If you're still uncertain on those, perhaps you could explicitly identify which steps are causing you trouble?)

What you can't do is say that the vector you picked at the beginning is one of the vectors in the kernel of $(T-\lambda_i I)$. Perhaps you weren't aware that you were implying that when you re-used the variable $v$ there? In that case you could also think of the mistake this way: In math you cannot re-use a variable to refer to something if it's already referring to something else.

And if you step back and look at the proof the way you presented it: Your $v$ was originally chosen knowing only that it was non-zero. The fact that you concluded that this rather arbitrary vector must be an eigenvector of $T$ looks suspicious - if your proof (as written) was correct then you would be able to pick any non-zero vector and show that it was an eigenvector! We know that that's not true, so we know there must have been a mistake somewhere in our proof.