A real normal endomorphism is orthogonally block-diagonalisable with 1x1 real blocks and 2x2 rotation-scaling blocks

Apply [L5] to a basis of $V$ to choose an orthonormal basis, and let $A\in M_n(\mathbb{R})$ be the matrix of $T$. By [L1], $A{A}^{\mathsf{T}}=A^{\mathsf{T}}A$. Since $A$ has real entries, its conjugate transpose over $\mathbb{C}$ is $A^{\mathsf{T}}$, so $A$ is normal on ${\mathbb{C}}^n$ with the standard Hermitian inner product from [L3]. Thus [L2] gives an orthonormal eigenbasis $(z_1,\dots ,z_n)$, say $A{z}_j={\mu }_j{z}_j$. If $n=0$, all subsequent lists are empty and give the asserted empty block decomposition.

For each eigenvalue $\lambda$, write $E_{\lambda }={\ker }_{\mathbb{C}}(A-\lambda I)$. Expanding $x={\sum }_j{c}_j{z}_j$ gives $(A-\lambda I)x={\sum }_j({\mu }_j-\lambda )c_j{z}_j$, so $E_{\lambda }$ is exactly the span of the $z_j$ with ${\mu }_j=\lambda$. Consequently these vectors form an orthonormal basis of the entire eigenspace, the distinct eigenspaces are mutually orthogonal, and their sum is ${\mathbb{C}}^n$. Also $A\bar{x}=\overline{Ax}$ by reality of $A$, so conjugation maps $E_{\lambda }$ onto $E_{\bar{\lambda }}$, with itself as inverse.

For real $\lambda$, put $R_{\lambda }={\ker }_{\mathbb{R}}(A-\lambda I)$. If $x=u+iw\in E_{\lambda }$ with $u,w\in {\mathbb{R}}^n$, comparison of real and imaginary parts gives $u,w\in R_{\lambda }$; conversely every such $u+iw$ lies in $E_{\lambda }$. Choose a real basis of the finite-dimensional subspace $R_{\lambda }$ and apply [L5] to obtain a real orthonormal basis. It also spans $E_{\lambda }$ over $\mathbb{C}$ by the decomposition just proved, and is complex-linearly independent: the real and imaginary parts of a complex linear relation are real relations among that real basis. Thus it is a full orthonormal complex basis of $E_{\lambda }$, consisting of real eigenvectors.

For each conjugate pair of nonreal eigenvalues, select the representative $\lambda =a+ib$ with $b>0$. Choose an orthonormal basis $v_1,\dots ,v_m$ of $E_{\lambda }$ from step 2.1 and use ${\bar{v}}_1,\dots ,{\bar{v}}_m$ for $E_{\bar{\lambda }}$. Conjugation is bijective between these eigenspaces and satisfies $⟨{\bar{v}}_j,{\bar{v}}_k⟩=\overline{⟨v_j,v_k⟩}={\delta }_{jk}$, so this is a full orthonormal basis there. Step 2.1 also gives $⟨v_j,{\bar{v}}_k⟩=0$ for every $j,k$.

Write $v_j=u_j+i{w}_j$ with real $u_j,w_j$. Using the coordinate pairing, the identities $⟨v_j,v_k⟩={\delta }_{jk}$ and $⟨v_j,{\bar{v}}_k⟩=0$ give respectively $u_j\cdot u_k+w_j\cdot w_k={\delta }_{jk}$, $w_j\cdot u_k-u_j\cdot w_k=0$, $u_j\cdot u_k-w_j\cdot w_k=0$, and $w_j\cdot u_k+u_j\cdot w_k=0$. Hence $u_j\cdot u_k=w_j\cdot w_k={\delta }_{jk}\mathrm{/}2$ and $u_j\cdot w_k=w_j\cdot u_k=0$. Therefore $e_j=\sqrt{2}{u}_j$ and $f_j=\sqrt{2}{w}_j$ form an orthonormal real list. Comparing parts in $A(u_j+i{w}_j)=(a+ib)(u_j+i{w}_j)$ gives $A{e}_j=a{e}_j-b{f}_j$ and $A{f}_j=b{e}_j+a{f}_j$. Thus each ordered pair $(e_j,f_j)$ spans an invariant plane with matrix $\left(\begin{array}{cc}a& b\\ -b& a\end{array}\right)$. Moreover $u_j=(v_j+{\bar{v}}_j)\mathrm{/}2$ and $w_j=(v_j-{\bar{v}}_j)\mathrm{/}(2i)$, so these real vectors have complex span exactly $E_{\lambda }\oplus E_{\bar{\lambda }}$.

Collect the real bases from step 3.1 and the real pairs from step 4.1 for every selected nonreal representative. Vectors from different groups are orthogonal: each group lies in the complex span of its own eigenspace or conjugate eigenspace pair, and these disjoint groups of eigenspaces are mutually orthogonal by step 2.1; for real vectors the Hermitian pairing is the real dot product. Within each group orthonormality was proved in steps 3.1 and 4.1. Their complex span is all of ${\mathbb{C}}^n$ by step 2.1. Since the collected vectors are real, taking real parts of a complex expansion of any $x\in {\mathbb{R}}^n$ shows that they span ${\mathbb{R}}^n$ over $\mathbb{R}$. They therefore give a real orthonormal basis, including the empty basis when $n=0$. Each basis vector from step 3.1 contributes $[\lambda ]$, and each pair from step 4.1 contributes the displayed block with $b>0$, hence $b\mathrm{\ne }0$. Transporting this basis through the coordinates of step 1.1 gives the required orthonormal basis of $V$.