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

"Lemma. let $\displaystyle T$ be a self-adjoint operator on a finite-dimensional inner product spave $\displaystyle V$. Then

(a) Every eigenvalue of $\displaystyle T$ is real.

(b) Suppose that $\displaystyle V$ is a real inner product space. Then the characteristic polynomial of $\displaystyle T$ splits."

The proof for part (b) is as follows:

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

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