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Sorry for my bad english.

Let $u : \mathbb{R} \rightarrow \mathbb{R}^n$ a solution of a linear differential equation $ x'(t) = Ax(t)$ $(1)$ with $ A \in M_{nn}(\mathbb{R})$. We suppose A is a diagonalizable matrix.

We want to prove that if this equation has a periodic solution (non-zero) with period $T > 0$, then the matrix $e^{TA}$ has the eigenvalue $1$ ; and to deduce that if all the eigenvalues of A are real, then $(1)$ hasn't periodic solution non constant.

I don't see how to make explicit it. Someone could help me ? Thank you in advance...

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  • $\begingroup$ Hint: in the Eigenvector basis, the equations simplify to $x'_i(t)=\lambda_i x_i(t)$ for which the solutions are obvious. $\endgroup$ – Yves Daoust May 1 '17 at 19:51
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If $A$ is diagonalizable then, there is a frame of reference in which the solved system is decoupled: $$y_i'(t) = \lambda_i\, y_i(t)$$

Being $A=P\Lambda P^{-1}$ and $\Lambda = diag{(\lambda_i)}$

This system of equations leads to the following solution: $$y(t) = \exp{(\Lambda t)} c$$ For any constant vector $c$

If the eigenvalues of $A$ (the same as the eigenvalues of the diagonalized matrix $\Lambda$) $\textit{i.e.}$ the numbers $\lambda_i$ are all complex the solutions are periodic, with period $T$:

$$y(t) = \exp{(\Lambda t)} c= \exp{(i\,\mathcal{Im}\left\{\Lambda\right\} t)} c$$ (note that $\exp{it}=\cos{t}+i\sin{t}$)

On the contrary they are not periodic, leading to exponentials that increase or decrease with time $$y(t) = \exp{(\Lambda t)} c$$

Recovering the original variable $x(t)$ one has $$x(t) = P\exp{(\Lambda t)} c$$

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