Splitting of the characteristic polynomial of a self-adjoint operator
From Linear Algebra by Friedberg, Insel, and Spence, 3rd edition:
"Lemma. let be a self-adjoint operator on a finite-dimensional inner product spave . Then
(a) Every eigenvalue of is real.
(b) Suppose that is a real inner product space. Then the characteristic polynomial of splits."
The proof for part (b) is as follows:
"Let , be an orthonormal basis for , and . Then is self-adjoint. Define by . Then is a linear operator on . Furthermore, is self-adjoint because , where is the standard ordered (orthonormal) basis for . So by (a) the eigenvalues of are real. By the fundamental theorem of algebra the characteristic polynomial of splits into factors of the form . Since each is real, the characteristic polynomial splits over . But has the same characteristic polynomial as , which has the same characteristic polynomial as . Therefore, the characteristic polynomial of splits."
I can follow the proof, but I don't understand why all the work in is necessary. Why doesn't the splitting of the characteristic polynomial of just follow straight from the fact that every eigenvalue of is real? The eigenvalues are the roots of the characteristic polynomial; if they're real, then shouldn't split over ?