Given an $n \times n$ normal matrix $A$ over $\mathbb{R}$, show that there is an orthogonal matrix $O$ such that $O^TAO$ is either diagonal or in the form
$$\left[\begin{matrix}{\lambda}&0&0\\0&a&b\\0&-b&a\end{matrix}\right].$$
So far, my thoughts are as follows. The characteristic polynomial $f$ of $A$ is cubic, so it has at least one real root, call it $\lambda$ with corresponding eigenvector $V_\lambda$, and $f = (x-\lambda)q$ for some polynomial $q$ of degree 2.
If $q$ has distinct real roots, then $A$ has an orthonormal basis of eigenvectors, which give the columns of a matrix $O$ such that $O^{-1}AO = O^TAO$ is diagonal.
Otherwise, ...? Essentially all of the theorems and exercises in my text work over $\mathbb{C}$, not $\mathbb{R}$. If we were working over $\mathbb{C}$, we would always have an orthonormal basis of eigenvectors $A$ would always be diagonalizable by such a matrix $O$.
You need to work out the details that the $2$-dimensional invariant subspace corresponding to eigenvalues $\lambda,\bar\lambda$ for a complex (non-real) eigenvalue $\lambda$, with the appropriate basis, gives the matrix $\begin{bmatrix} a&b\\-b&a\end{bmatrix}$ when $\lambda = a+ib$. (Hint: If $v\in\Bbb C^3$ is an eigenvector with eigenvalue $\lambda$ of the complexification of $A$, write $v=x+iy$ for $x,y\in\Bbb R^3$.)