System of first order linear ODEs (real case)

It is a particular case of the linear complex case and, more in general, of the system of first order ODEs.
If $A$ is a real matrix, given a solution $\phi (t)$, the conjugate function is just the solution of the system whose initial value is the conjugate of that of $\phi (t)$.
By uniqueness of solutions, if we take any real initial condition, $v\in {\mathbb{R}}^n$ then

must be real, and so $e^{At}$ is a real matrix.

What form does the elements of this matrix have? In the note of the complex case we stated that

Solutions are then made by mixing quasi-polynomials by mean of complex number, so the results are linear combinations of real and imaginary parts of these quasi-polynomials:

Proposition

Let $\dot{z}=Az$ a system of first order linear equations, with $A:{\mathbb{R}}^n\mapsto {\mathbb{R}}^n$ a linear operator. Then, if $A$ has real eigenvalues ${\lambda }_i$ with algebraic multiplicities $n_i$ and complex eigenvalues ${\alpha }_j\pm i{\omega }_j$ with multiplicities ${\mu }_j$ then every component of a solution $\phi (t)$ has the form:

where $p_{m,i}(t)$, $q_{m,j}(t)$ and $r_{m,j}(t)$ are real polynomials with degree limited by the multiplicities.
$◼$
In particular, consider the case where all eigenvalues are single. We would have:

with $a,b,c$ families of real constants. But there are too many.

How many of degrees of freedom do we have? Let's see.
The map $\phi :I\subset \mathbb{R}\mapsto {\mathbb{C}}^n$ is a complex solution of the equation if and only if the real and the imaginary part of $z$ are both real solutions. This is because of linearity and because $\bar{z}(t)$ is also a solution of the associated complex equation (the only one with initial value the complex conjugate of $z(0)$).

Suppose $A$ has $n$ distinct eigenvalues, $r$ real eigenvalues and $c$ complex eigenvalues, such that $n=r+2c$. The general solution is

where we have chosen real eigenvectors for the real eigenvalues, and complex conjugate eigenvectors for the complex eigenvalue pairs. We observe that for $\phi (0)$ being real

and therefore coefficients of complex conjugates eigenvectors must be complex conjugates. So, for real eigenvalues (with real eigenvector) $c_k$ is real, and we denote it by $a_k$. And for the complex conjugates eigenvalues ${\lambda }_k$ and ${\lambda }_j$, with of course complex conjugates eigenvectors, coefficients $c_k$ and $c_j$ must be conjugates. So we can write the solution as

and so

In conclusion, in equation (1) above, $c_{m,i}$, $a_{m,j}$, $b_{m,j}$ are "connected" in such a way that they only suppose $n$ degrees of freedom: they are linear functions of real and imaginary parts of $c_k$ and we have that the first $r$ $c_k$s are real and the last $2c$ are paired by conjugation .
Sometimes it is preferred other expression of the general solution. For $k=r+1..r+c$ the terms above can be rewritten as:

And this can be cleaned be means of a change of names and trigonometric properties into:

where $v_k^1=Re({\xi }_k)$, $v_k^2=Im({\xi }_k)$, $A_k=|2c_k|$ and ${\varphi }_k=arg(2c_k))$