Prior to reading the following post please consider the following notation.
- $T\in\mathcal{L}(V)$ means that $T$ belongs to the set of linear operators on the vector space $V$.
- Given an operator $T\in\mathcal{L}(V)$ and a polynomial $p(z) = c_0+c_1z+c_2z^2+\cdot\cdot\cdot+c_mz^m$. Then $p(T)$ denotes the operator $p(T) = c_0I+c_1T+c_2T^2+\cdot\cdot\cdot+c_mT^m$
In Linear Algebra Done Right Axler presents an argument to prove the following theorem
Theorem. Every operator over a non-zero finite dimensional complex vector space $V$ has an eigenvalue.
He later asks the reader to re-write the argument using the linear map that sends $p\in\mathcal{P}_n(\mathbf{C})$ to $p(T)v$.
The following is my attempt at re-writing the proof. Is it Correct? In the following Proof i make use of the well known theorem.
($4.14$) If $p\in\mathcal{P}(\mathbf{C})$ is a nonconstant polynomial, then $p$ has a unique factorization (except for the order of the factors) of the form $p(z) = c(z-\lambda_1)\cdot\cdot\cdot(z-\lambda_m)$ where $c,\lambda_1,...,\lambda_n\in\mathbf{C}$.
Proof. Let $V$ be a finite dimensional complex vector space such that $\dim V = n>0$ and let $T$ be an arbitrary linear operator on $V$. Since $V$ is non-zero we may invoke the existence of a vector $v\in V$ such that $v\neq 0$.
Consider now the map $\omega\in\mathcal{L}(\mathcal{P}_n(\mathbf{C}),V)$ defined such that $\omega(p) = p(T)v$, evidently $\dim\mathcal{P}_n(\mathbf{C})) = n+1$ indicating that $\omega$ is not injective, equivalently we have some non-zero polynomial $q = \alpha_0+\alpha_1z+\alpha_2z^2+\cdot\cdot\cdot+\alpha_mz^m$ such that $\omega(q) = q(T)v = 0$ furthermore since $q$ is non-zero it follows that not all $\alpha_0,\alpha_1,\alpha_2,\ .\ .\ .\ ,\alpha_m$ equal $0$ and in particular not all of the $\alpha_1,\alpha_2,\ .\ .\ . ,\alpha_m$ are $0$ since that would imply that $q(T)v = \alpha_0I = \alpha_0v = 0$ and by extension $\alpha_0 = 0$ resulting in a contradiction.
Having established that at least one of $\alpha_1,\alpha_2,\ .\ .\ .\ ,\alpha_m$ is not zero it follows that $\deg q\ge 1$ and thus by theorem $\textbf{4.14}$ we may factorize the operator $q(T)$ to yield $q(T) = \beta(T-\lambda_1I)(T-\lambda_2I)\cdot\cdot\cdot(T-\lambda_kI)$ where $\beta\neq 0$ consequently $\beta(T-\lambda_1I)(T-\lambda_2I)\cdot\cdot\cdot(T-\lambda_kI)v = 0$ implying that at least one of the operators $(T-\lambda_1I),(T-\lambda_2I),\ .\ .\ .\ ,(T-\lambda_kI)$ is not injective therefore at least of $\alpha_1,\alpha_2,\ .\ .\ .\ ,\alpha_k$ must be an eigenvalue of $T$.
$\blacksquare$
PS: Are there any proofs of the above theorem that do not make use of theorem $4.14$.