From Linear Algebra by Friedberg, Insel, and Spence, 3rd edition:

"Lemma. let T be a self-adjoint operator on a finite-dimensional inner product spave V. Then
(a) Every eigenvalue of T is real.
(b) Suppose that V is a real inner product space. Then the characteristic polynomial of T splits."

The proof for part (b) is as follows:
"Let n=\mbox{dim}(V), \beta be an orthonormal basis for V, and A=[T]_\beta. Then A is self-adjoint. Define T_A: \mathbb{C}^n \to \mathbb{C}^n by T_A(x)=Ax. Then T_A is a linear operator on \mathbb{C}^n. Furthermore, T_A is self-adjoint because [T_A]_\gamma=A, where \gamma is the standard ordered (orthonormal) basis for \mathbb{C}^n. So by (a) the eigenvalues of T_A are real. By the fundamental theorem of algebra the characteristic polynomial of T_A splits into factors of the form x-\lambda. Since each \lambda is real, the characteristic polynomial splits over \mathbb{R}. But T_A has the same characteristic polynomial as A, which has the same characteristic polynomial as T. Therefore, the characteristic polynomial of T splits."

I can follow the proof, but I don't understand why all the work in \mathbb{C}^n is necessary. Why doesn't the splitting of the characteristic polynomial of T just follow straight from the fact that every eigenvalue of T is real? The eigenvalues are the roots of the characteristic polynomial; if they're real, then shouldn't T split over \mathbb{R}?