Here is some quick note for the existence of eigenvector for linear operator $latex T$ in complex vector space of dimension $latex n$. Consider any non-zero vector $latex X$ in the space, we have the $latex n+1$ vectors $latex X, T[X],T^2[X],T^3[X],\cdots,T^n[X]$ to be linearly dependent. Thus, there exists $latex a_0,a_1,\cdots,a_n$ such that $latex a_0+a_1T[X]+a_2T^2[X]+\cdots+a_nT^n[X]=0$. If $latex a_n$ is zero, we will pick the largest $latex k$, $latex k\le n$, such that $latex a_k \neq 0$. In any case, we can write
$latex a_0+a_1T[X]+a_2T^2[X]+\cdots+a_kT^k[X]=0$ with $latex a_k\neq 0$.
Now consider the polynomial expression
$latex a_0+a_1x+a_2x^2+\cdots+a_kx^k$. We can find complex numbers $latex c_1,c_2,\cdots,c_k$ such that
$latex a_0+a_1x+a_2x^2+\cdots+a_kx^k=a_k (x-c_1)(x-c_2)\cdots(x-c_k)$. Easy to verify that
$latex a_k(T-c_1 I)(T-c_2 I)\cdots(T-c_k I)[X]$
$latex =a_0+a_1T[X]+a_2T^2[X]+\cdots+a_kT^k[X]=0$.
For the above to be true, there must exist an $latex l$, $latex 1\le l \le k$, such that
$latex (T-c_{l} I)\cdots(T-c_kI)[X]=0$ but $latex (T-c_{l+1} I)\cdots(T-c_k I)[X]\neq 0$. Define $latex Y_l=(T-c_{l+1} I)\cdots(T-c_k I)[X]$, note that $latex (T-c_{l} I )[Y_l]=0$ with $latex Y_l\neq 0$. That is $latex T[Y_l]=c_l Y_l$ and thus $latex c_l$ and $latex Y_l$ are precisely an eigenvalue and eigenvector of $latex T[\cdot]$.