Necessar and sufficient condition of similitude Here is the problem : 

Give a necessar and sufficient over $M\in \mathcal{M}_n(\mathbb{C})$ (a complex matrix) such that there exists $P \in GL_n(\mathbb{C})$ and $N \in \mathcal{M}_n(\mathbb{R})$ satisfying : $$M=P^{-1}NP$$

I have barely no ideas how to start. The problem is difficult and has been proposed to a french oral examination (ENS)
If you have the solution to this problem or hints please give them. 
Thank you. 
 A: A real subspace $V \subseteq \mathbb{C}^n$ is called totally real if $V \cap iV = \{0\}$. Let $W := \mathrm{span} \{e_1, \ldots, e_n \}$ where $e_i$ are the standard basis vectors of $\mathbb{C}^n$. Then $W$ is totally real and a complex matrix $M \in M_n(\mathbb{C})$ is in fact a real matrix if and only if $Me_j \in W$ for all $1 \leq j \leq n$. In other words, a matrix $M$ is real if and only if the totally real subspace $W$ is $A$-invariant.
A complex matrix $M \in M_n(\mathbb{C})$ will be similar (using an invertible complex matrix) to a real matrix if and only if there exists a $A$-invariant totally-real subspace $V$ of real dimension $n$.

Notation: Given a complex matrix $A \in M_n(\mathbb{C})$, we denote by $T_A \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ the corresponding linear map $T_A(w) = Aw$.
Assume that such a subspace exists and choose a real basis $v_1, \ldots, v_n$ for $V$. Then $v_1, \ldots, v_n, iv_1, \ldots, iv_n$ will be a real basis for $\mathbb{C}^n$. Define a real linear map $T \colon \mathbb{C}^n \rightarrow \mathbb{C}^n$ by requiring that $T(v_j) = e_j$ and $T(iv_j) = ie_j$. One can easily check that $T$ is in fact $\mathbb{C}$-linear and invertible and thus represented by a complex invertible matrix $P$ (so $T = T_P$). We have $T_P(V) = W$ and $T_P(iV) = iW$ and so $T_P$ transforms the direct sum decomposition $\mathbb{C}^n = V \oplus iV$ to the direct sum decomposition $\mathbb{C}^n = W \oplus iW$. We have
$$ (T_P \circ T_M \circ T_P^{-1})(e_j) = T_P(T_M(v_j)) \in T_P(T_M(V)) \subseteq T_P(V) = W $$
and thus $PMP^{-1}$ is a real matrix.
In the other direction, if $PMP^{-1} = N$ is real, let $V = \mathrm{span} \{ v_1 := P^{-1}e_1, \ldots, v_n := P^{-1}e_n \}$. Then
$$ T_M(v_j) = T_M(T_P^{-1}(e_j)) = T_P^{-1} T_N(e_j) \in T_P^{-1}(W) = V $$
so $V$ is a real $n$-dimensional $M$-invariant subspace. I'll leave it to you to verify that $V$ is also totally real.
A: You can start with the real Jordan form. 
The following fact is useful: given two $k\times k$ Jordan blocks $J_k(\lambda)$ and $J_k(\bar{\lambda})$, the matrix $J_k(\lambda)\oplus J_k(\bar{\lambda})$ is similar to the matrix
$$\tag{1}
\pmatrix{K(\lambda)&I_2\\&K(\lambda)&I_2\\&&\ddots&\ddots\\&&&K(\lambda)&I_2\\&&&&K(\lambda)\\}\in\mathbb{R}^{2k\times 2k},
\quad
K(\lambda)=\pmatrix{\Re(\lambda)&\Im(\lambda)\\-\Im(\lambda)&\Re(\lambda)},
$$
where $I_2$ is the $2\times 2$ identity matrix.
Consequently, every real matrix has a real Jordan form where the blocks associated with nonreal eigenvalues have the form (1). In addition, the similarity transformation matrix is real.
We have hence that a complex matrix is similar to a real matrix if and only if it has a real Jordan form. That is, a complex matrix is similar to a real matrix if and only if for every $k=1,2,\ldots$ and every eigenvalue $\lambda$, the number of Jordan blocks $J_k(\lambda)$ and $J_k(\bar{\lambda})$ is equal, that is, each Jordan block associated with a nonreal eigenvalue $\lambda$ can be associated with a unique Jordan block of the same size corresponding to $\bar{\lambda}$.
A simpler criterion (following from the reasoning above) states that a complex matrix $A$ is similar to a real matrix if and only if $A$ is similar to $\bar{A}$.
