Proof that each square matrix has eigenvectors Suppose I have an $n \times n$  matrix $B$ over $\mathbb{C}^n$. How can I show that $B$ has at least one eigenvector, aka prove that for some $x$ we have $(A - \lambda I) x = 0$?
 A: Sketch:
Step 1: The characteristic polynomial of an $n\times n$ matrix $B$ is a polynomial of degree $n$.  By the fundamental theorem of algebra, this polynomial has at least one root, let $\lambda$ be a root of the characteristic polynomial.
Step 2: By construction, $\det(B-\lambda I)=0$.  This implies, by the invertible matrix theorem that $B-\lambda I$ is not invertible, $B-\lambda I$ is not full rank, etc.  In particular, if you row reduce the augmented matrix $\begin{bmatrix}B-\lambda I&|&\vec{0}\end{bmatrix}$, you will have a free variable.  By letting this free variable be something other than $0$, you can construct your eigenvector.
A: Choose any nonzero vector $x$ and consider the $n+1$ vectors $\{ x, Ax, A^2, \cdots, A^{n-1}x, A^n x\}$.
They are linearly dependent and hence there exist $\{a_0, a_1, \cdots, a_n\}$, not all zero, s.t.
$$a_0 x+a_1Ax+a_2A^2x+\cdots+a_nA^nx=0$$
Suppose only $a_0$ is nonzero, then
$$a_0 x =0 \implies x=0,$$
Constradiction.
So there exist largest integer $m\ge1$ s.t. $a_m\ne0$.
WLOG we can let $a_m=1$
By fundamental theorem of algebra, we have
$$a_0I+a_1A+\cdots+a_mA^m=(A-b_m I)(A-b_{m-1}I)\cdots(A-b_2I)(A-b_1I)$$
$$0=(A-b_mI_n)(A-b_{m-1}I)\cdots(A-b_2I)(A-b_1I)x$$
Let $k$ be the smaller integer s.t.
$$(A-b_kI)(A-b_{k-1}I)\cdots(A-b_1 I)x=0$$
Then by definition
$$z=(A-b_{k-1}I)\cdots(A-b_1 I)x\ne0$$
and
$$(A-b_kI)z=0$$
$$Az=b_k z$$
Q.E.D.
