Canonical form of normal operator in Euclidean space 
Let $V$ be a finite-dimensional Euclidean space and $f$ be a normal
  operator on $V$. Then there is an orthonormal basis of $V$ such that
  the matrix of $f$ in this basis is block-diagonal consisting of blocks
  of the size $1\times 1$ and blocks of the size $2\times 2$ of the form
  $\begin{pmatrix} \mu & -\nu \\ \nu & \mu \end{pmatrix}$. This matrix is called
  canonical form of normal operator and corresponding basis is called canonical basis.

I was able to prove this theorem by myself using my previous
 post. So let's discuss the algorithm how to find such canonical
 form:

Consider complexification $f_{\mathbb{C}}$ of our operator $f$.
  Operator $f_{\mathbb{C}}$ has the same matrix as $f$. Let
  $\lambda=\mu+i\nu$ be a root of characteristic polynomial of this
  matrix. We'll find the basis of corresponding eigenspace
  $V_{\lambda}\subset V$ and orthogonalize it (with respect to
  Hermitian inner product): let $z_1,\dots,z_l$ - desired orthogonal
  basis, where $x_j, y_j$ are real and imaginary parts of $z_j$:
  $z_j=x_j+iy_j$, $j=1,\dots,l$. 
Then the system of real vectors $\{x_1,y_1,x_2,y_2,\dots,x_l,y_l\}$ is
  orthogonal with $|x_j|=|y_j|$, $j=1,\dots,l$ (with respect to
  Euclidean inner product), and operator $f$ acts on them in the desired
  way: $f(x_j)=\mu x_j+\nu y_j, f(y_j)=-\nu x_j+\mu y_j$. Let's
  normalize vectors $$x'_j=\frac{x_j}{|x_j|}, \quad
 y'_j=\frac{y_j}{|y_j|}$$ and include them to the canonical basis: each
  such pair of vectors $x'_j,y'_j$ corresponds to the block
  $\begin{pmatrix} \mu & -\nu \\ \nu & \mu \end{pmatrix}$.

I was trying to understand some moments of the second excerpt but I failed, so let me clarify my questions:
Questions: 1) How to show that the system of real vectors $\{x_1,y_1,x_2,y_2,\dots,x_l,y_l\}$ is
 orthogonal with $|x_j|=|y_j|$, $j=1,\dots,l$ (with respect to
 Euclidean inner product)?
2) Since $f_{\mathbb{C}}(z_k)=\lambda z_k$ then $f(x_k)+if(y_k)=(\mu+i \nu)(x_k+iy_k)=(\mu x_k-\nu y_k)+i(\nu x_k+\mu y_k)$ it follows $f(x_k)=\mu x_k-\nu y_k$ and $f(y_k)=\nu x_k+\mu y_k$.  And we see that my answer differs from the above one a bit.  So I guess in this case it is more natural to consider the system $\{y_1,x_1,\dots,y_l,x_l\}$. Right?
3) Suppose that roots of characteristic polynomial are $\lambda_1,\dots,\lambda_k$ and $V_{\lambda_1},\dots,V_{\lambda_k}$ are corresponding eigenspaces. It could happen that $V_{\lambda_1}\oplus \dots \oplus V_{\lambda_k}\subsetneq V$. Using the above algorithm I can construct basis for each $V_{\lambda_i}$ but what I should do with the rest, since their direct sum is a proper subspace?
Would be very grateful if someone will answer my questions, please?
EDIT (Possible answer to question 1): I guess the algorithm above is applied for purely complex root, i.e. $\nu \neq 0$. We see that $Az_k=\lambda z_k$ then $A\overline{z_k}=\overline{\lambda}\overline{z_k}$ because $A$ is real matrix. Since $\nu \neq 0$ as I said then $z_k \perp \overline{z_k}$, i.e. $$0=\langle x_k+iy_k,x_k-iy_k\rangle=|x_k|^2-|y_k|^2+2i\langle y_k,x_k\rangle$$ which shows us that $\langle x_k,y_k\rangle=0$ and $|x_k|=|y_k|$.
And somehow I need to show that $x_i\perp x_j, y_i\perp y_j$ and $x_i\perp y_j$ for $i\neq j$. 
Let $k\neq l$ and I am going to show that $\langle x_k,x_l\rangle=\langle y_k,y_l\rangle=\langle x_k,x_l\rangle=0$. I will show just the first one. We see that $2x_k=z_k+\overline{z_k}$ and $2x_l=z_l+\overline{z_l}$ then it follows that $$\langle x_k,x_l\rangle=\frac{1}{4}\langle z_k+\overline{z_k},z_l+\overline{z_l}\rangle=\dfrac{1}{4}\left[\langle z_k,z_l\rangle+\langle z_k,\overline{z_l}\rangle+\langle\overline{z_k}, z_l\rangle+\langle\overline{z_k},\overline{z_l}\rangle\right].$$
We already know that $\langle z_k,z_l\rangle=0$. Since $Az_k=\lambda z_k$ then $A\overline{z_l}=\overline{\lambda}\overline{z_l}$ and because $\lambda \neq \overline{\lambda}$ and since $z_k$ and $\overline{z_l}$ are eigenvectors corresponding to different eigenvalues then it follows that $\langle z_k,\overline{z_l}\rangle=0$. In the same way one can show $\langle \overline{z_k},z_l\rangle=0.$
It is easy to show that $\langle\overline{z_k},\overline{z_l}\rangle=0$ and it follows because $\langle z_k,z_l\rangle=0$. More precisely, $$\langle\overline{z_k},\overline{z_l}\rangle=\operatorname{Re}\langle z_k,z_l\rangle -i\operatorname{Im}\langle z_k,z_l\rangle=0.$$
The same reasoning can be applied for the rest pairs.
 A: Regarding question 1: your proof is correct and seems to be the most reasonable approach.
Regarding question 2: I agree. You could also use blocks of the form $\pmatrix{\mu & \nu\\-\nu & \mu}$ instead.
Regarding question 3: The rest of the eigenvalues are real.  So, we simply take the remaining blocks to be $1 \times 1$.
A: Let $V$ be Euclidean space and $f$ be a normal operator on $V$. Consider complexification  $V_{\mathbb{C}}$ and $f_{\mathbb{C}}$ of $V$ and $f$, respectively. One can check that $f_{\mathbb{C}}$ normal operator on Hermitian space $V_{\mathbb{C}}$. Let $\text{Sp}f_{\mathbb{C}}=\{\lambda_1,\dots,\lambda_k,\dots,\lambda_m\}$ be eigenvalues of $f_{\mathbb{C}}$ and suppose the first $k$ are complex and the last $m-k$ are real numbers. Since $f_{\mathbb{C}}$ is a normal operator on Hermitian space $V_{\mathbb{C}}$ then it is diagonalizable, i.e. $$V_{\mathbb{C}}=\bigoplus \limits_{j=1}^m V^{\mathbb{C}}_{\lambda_j},$$ where by $V^{\mathbb{C}}_{\lambda_j}$ I mean eigenspace corresponding to $\lambda_j$, i.e. $V^{\mathbb{C}}_{\lambda_j}=\ker (f_{\mathbb{C}}-\lambda_j\cdot \text{id})$.
Remark: Since $V_{\mathbb{C}}$ is $V\oplus V$ equipped with operator of complex structure $J(u,v)=(-v,u)$ then for my convenience I will denote elements of $V_{\mathbb{C}}$ as pair of elements from $V$.
1. If $\lambda\in \mathbb{R}$ then $V^{\mathbb{C}}_{\lambda}$ has "real" basis say $\{(v_1,0),\dots,(v_n,0)\}$.
Indeed, consider $V_{\lambda}=\ker(f-\lambda\cdot \text{id})$ and its easy to see that $\dim V_{\lambda}=\dim V^{\mathbb{C}}_{\lambda}$ because $A_{f_{\mathbb{C}}}=A_f$ in some basis. Let $\{v_1,\dots,v_n\}$ is a basis for $V_{\lambda}$ then one can show that $\{(v_1,0),\dots,(v_n,0)\}$ is a basis for $V^{\mathbb{C}}_{\lambda}$.
2. Let $\lambda=\mu+i\nu$ with $\nu\neq 0$ then $\lambda\neq \overline{\lambda}$. Also easy to see that $\dim V^{\mathbb{C}}_{\lambda}=\dim V^{\mathbb{C}}_{\overline{\lambda}}=n$ (this $n$ is not the same as the previous one). Then $\dim V^{\mathbb{C}}_{\lambda}\oplus V^{\mathbb{C}}_{\overline{\lambda}}=2n$. Let $\{z_1,\dots,z_n\}$ is orthogonal basis of $V^{\mathbb{C}}_{\lambda}$, where $z_j=(x_j,y_j)$. Then by reasoning which have demonstrated in my question one can show that $(x_1,0),(0,y_1),\dots,(x_n,0),(0,y_n)$ is orthogonal and $|x_j|=|y_j|$ and since $z_j$ is basis then it follows that those $2n$ vectors are linearly independent and forms the basis of $V^{\mathbb{C}}_{\lambda}\oplus V^{\mathbb{C}}_{\overline{\lambda}}$. It is easy exercise to show that $(x_1,0),(y_1,0),\dots,(x_n,0),(y_n,0)$ is still a basis (I just changed $(0,y)$ to $(y,0))$. 
Since each complex eigenvalue of $f_\mathbb{C}$ appears in pair (with its conjugate) then we can do that for each such pair. In general we will get a basis for $V_{\mathbb{C}}$ of the form $\{(e'_i,0)\}_{i=1}^n$, i.e. "real" basis of $V_{\mathbb{C}}$. Then this is also an easy exercise to show that $\{e'_i\}_{i=1}^n$ is basis for $V$.
P.S. May be I was to picky in the proof but I just want to know that I am understanding all this correctly. Thanks a lot for your attention, Omnomnomnom!
