Solutions to Linear System of ODEs Have Identical Bases?

When solving an ODE system like $\vec{x}'=\begin{bmatrix} 1 & 2\\3&4 \end{bmatrix}\vec{x},$ you assume that the solutions $x_1(t)$ and $x_2(t)$ will have identical bases. So in this example, each solution might be a linear combination of two exponentials, $e^{r_1t}$ and $e^{r_2t}.$ This allows the system to be solved as an eigenvalue equation. My question is, why can we guess that $x_1$ and $x_2$ will have the same basis functions? Why doesn't the solution look more like this: $x_1=c_1e^{r_1t}+c_2e^{r_2t}$ and $x_2=c_3e^{r_3t}+c_4e^{r_4t}?$

I think you can benefit from reading this, but in short the solutions to the equation $\dot{x} = A x$ are in general $x(t) = e^{At}x(0)$, so if you can diagonalize $A = U\Lambda U^{-1}$ then $x(t) = U e^{\Lambda t}U^{-1}x_0$, where $e^{\Lambda t} = {\rm diag}\{e^{\lambda_1 t}, e^{\lambda_2t}, \cdots \}$, so in general the solution will be a linear combination of functions of the form $e^{\lambda t}$. What is important to note here is that in general $\lambda \in \Bbb{C}$, so solutions can also include functions such as $\sin \lambda t$

EDIT

Note that

\begin{eqnarray} A^2 &=& (U\Lambda U^{-1})(U\Lambda U^{-1}) = U\Lambda(UU^{-1})\Lambda U^{-1} = U\Lambda^2U^{-1} \\ &\vdots& \\ A^k &=& (U \Lambda^{k-1} U^{-1})(U\Lambda U^{-1}) = U \Lambda^k U^{-1} \end{eqnarray}

Therefore

\begin{eqnarray} e^{At}&=& \sum_{k=0}^{+\infty}\frac{t^k}{k!}A^k = \sum_{k=0}^{+\infty}\frac{t^k}{k!}U\Lambda^kU^{-1} \\ &=& U\left(\sum_{k=0}^{+\infty}\frac{t^k}{k!}\Lambda^k\right)U^{-1} \\ &=& U e^{\Lambda t}U^{-1} \end{eqnarray}

• Thank you for the answer. Why is it that $e^{At} = Ue^{\Lambda t}U^{-1}$? How come the matrices $U$ and $U^{-1}$ aren't powers of an exponential? – Aladris May 16 '17 at 14:09
• @Aladris I updated my answer, hopefully it makes more sense now – caverac May 16 '17 at 14:18