## Splitting of the characteristic polynomial of a self-adjoint operator

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 $\mathbb{R}$?