Normal operator over real inner product space Let $L: V \rightarrow V$ be a linear operator on a finite dimensional real inner product space $V$ such that $L^{*} = L^{3}$. Show that $L^{2}$ is diagonalizable over $\mathbb{R}$.
Attempt: Suppose $L^{*} = L^{3}$. Then $LL^{*} = LL^{3} = L^{3}L = L^{*}L$. It follows that $L$ is a normal. By the real spectral theorem (I am only used to the complex version), there exists an orthonormal basis of $V$ consisting of eigenvectors of $L$, say $\{v_{1},..., v_{n} \}$. But then this consists of eigenvectors of $L^{2}$ since $Lv_{j} = \lambda_{j}v_{j} \implies L^{2}v_{j} = \lambda_{j}^{2}v_{j}$ and so $L^{2}$ is diagonalizable. 
Is my use of the spectral theorem legal?
 A: You have the right basic idea, but you have yet to prove the conclusion. We can extend the real vector space $V$ into a complex inner product space $\tilde V$.  $L$ is unitarily diagonalizable over $\tilde V$ by the spectral theorem, which means that $L^2$ is also unitarily diagonalizable over $\tilde V$ (since it has the same eigenvectors).
In order to conclude that $L^2$ is diagonalizable over the real vector space $V$, we must first conclude that its eigenvalues are real.  In order to do so, note that since $L^* = L^3$ and $L$ is (untiarily!!) diagonalizable, each eigenvalue $\lambda$ of $L$ satisfies $\overline{\lambda} = \lambda^3$, and therefore conclude that the only possible eigenvalues of $L$ are...?
Alternatively: note that $(L^2)^2 = L^*L$ necessarily has non-negative eigenvalues, from which we may conclude that $L^2$ has real eigenvalues.
From there, it suffices to note that for a real $\mu$, $\ker (L^2 - \mu I)$ has the same dimension as a real subspace of $V$ as it does as a complex subspace of $\tilde V$.
A: I am not sure the use of the spectral theorem for L is correct. For $L$ normal it states that: $L=ULU^*$ where $U$ is unitary, i.e. complex, this cannot be restricted to $\mathbb{R}$. 
An example of such a linear operator would be:
 $[{\begin{array}{cc}
   2 & -3 \\ 3 & 2 \      \end{array} }]
$. It is a real normal matrix but cannot be diagonalized into real values.
Instead, the approach I see is to realize $M=LL^*=L^4$ is symmetric. Applyting the spectral theorem yields it can be diagonalized into real eigenvectors and eigenvalues. Besides, being an even power the matrix will be positive semidefinite. This means that:
$ LL^* = L^4 = SDS^{-1} $
$L^2 = S \sqrt{D} S^{-1} $
