# Existence of Eigenvector in Linear Operator in Complex Vector Space

Here is some quick note for the existence of eigenvector for linear operator $T$  in complex vector space of dimension $n$. Consider any non-zero vector $X$ in the space, we have the $n+1$ vectors $X, T[X],T^2[X],T^3[X],\cdots,T^n[X]$ to be linearly dependent. Thus, there exists $a_0,a_1,\cdots,a_n$ such that $a_0+a_1T[X]+a_2T^2[X]+\cdots+a_nT^n[X]=0$. If $a_n$ is zero, we will pick the largest $k$, $k\le n$, such that $a_k \neq 0$. In any case, we can write
$a_0+a_1T[X]+a_2T^2[X]+\cdots+a_kT^k[X]=0$ with $a_k\neq 0$.
Now consider the polynomial expression
$a_0+a_1x+a_2x^2+\cdots+a_kx^k$. We can find complex numbers $c_1,c_2,\cdots,c_k$ such that

$a_0+a_1x+a_2x^2+\cdots+a_kx^k=a_k (x-c_1)(x-c_2)\cdots(x-c_k)$. Easy to verify that

$a_k(T-c_1 I)(T-c_2 I)\cdots(T-c_k I)[X]$
$=a_0+a_1T[X]+a_2T^2[X]+\cdots+a_kT^k[X]=0$.

For the above to be true, there must exist an $l$, $1\le l \le k$, such that

$(T-c_{l} I)\cdots(T-c_kI)[X]=0$ but $(T-c_{l+1} I)\cdots(T-c_k I)[X]\neq 0$.  Define $Y_l=(T-c_{l+1} I)\cdots(T-c_k I)[X]$, note that $(T-c_{l} I )[Y_l]=0$ with $Y_l\neq 0$. That is $T[Y_l]=c_l Y_l$ and thus $c_l$ and $Y_l$ are precisely an eigenvalue and eigenvector of $T[\cdot]$.