T normal operator has splitting characteristic polynomial - eigenvectors form basis?

Question: $\displaystyle T$ is a normal operator on a finite-dimensional real inner product space $\displaystyle V$ with a characteristic polynomial that splits. Prove that $\displaystyle V$ has an orthonormal basis of eigenvectors of $\displaystyle T$.

By Schur's theorem, because the characteristic polynomial of $\displaystyle T$ splits, there exists an orthonormal basis $\displaystyle \beta$ for $\displaystyle V$ such that $\displaystyle [T]_{\beta}$ is upper triangular. How does the fact that the inner product space being real ensure that such a basis is composed of the eigenvectors of $\displaystyle T$? Thanks!