Explain why arbitrary linear operator $T: \mathbb{C}^n \to \mathbb{C}^n$ has at least one eigenvector in $\mathbb{C}^n$. The way I'm thinking about it is if $T$ has at least one eigenvalue in $\mathbb{C}^n$, then there must be at least one eigenvector belonging to that eigenvalue. The problem is proving that there has to be at least one eigenvalue.

Because it has at least one eigenvalue. The characteristic polynomial must have at least one distinct root in $\mathbb{C}$. Unlike $\mathbb{R}$. 
