Show every Linear Transformation $T:V \to V$ has an eigenvector Let $V$ be a finite dimensional complex vector space.
How do we prove that any $T:V \to V$ must have an eigenvector for some eigenvalue. I understand that this would not be the case if $V$ was a Real Vector space but I am not sure how to prove it in either case.
Thanks
 A: Let $n = \dim V$.
Choose $v \in V$ with $v \ne 0$.  Then
$$
v, Tv,T^2 v, \dots , T^n v
$$
is not linearly independent, because $V$ has dimension $n$ and we have
$n+1$  vectors.  Thus there exist complex numbers $a_0, \dots , a_n$, not all
$0$,  such  that
$$
0 = a_0 v + a_1 Tv + \dots + a_n T^n v.
$$
Note that $a_1, \dots, a_n$ cannot all be $0$, because otherwise the equation above would become $0 = a_0 v$, which would force $a_0$ also to be $0$.
Make the $a$'s the coefficients of a polynomial, which by the Fundamental Theorem of Algebra has a factorization
$$
a_0 + a_1 z + \dots + a_n z^n = c (z - \lambda_1) \dotsm (z - \lambda_m),
$$
where $c$ is a nonzero complex number, each $\lambda_j \in \mathbf{C}$,  and
the equation holds for all $z \in \mathbf{C}$ (here $m$ is not necessarily equal to $n$, because $a_n$ may equal $0$).  We then have
\begin{align*}
0 &= a_0 v + a_1 Tv + \dots + a_n T^n v \\
&=(a_0 I + a_1 T + \dots + a_n T^n) v\\
&= c(T - \lambda_1 I) \dotsm (T - \lambda_m I)v.
\end{align*}
Thus $T - \lambda_j I$ is not injective for at least one $j$.  In
other words, $T$ has an eigenvalue.
