Logic behind solving a system of linear ODEs So if I have a system of linear ODES, I can write this in matrix form $x'=Ax$. I can solve this by determining the eigenvalues and eigenvectors of $A$. My question is why is this the case? Why is it that the eigenvalue is the coefficient in the exponential and why is it that we get the eigenvectors? Is there a way of showing this must always be the case?
 A: Suppose $A$ is diagonalizable, i.e., that we can write $A = SDS^{-1}$, where $D$ is a diagonal matrix whose diagonal entries consist of the eigenvalues of $A$ (denoted $\lambda_i$), and where $S$ is a matrix whose columns are the eigenvectors of $A$ (denoted $v_i$). This can always be done, just as long as $A$ has $n$ linearly independent eigenvectors (where $n$ is the dimension of $A$). We have
$$x' = Ax = SDS^{-1}x$$
$$\implies S^{-1}x' = (S^{-1}x)'= DS^{-1}x$$
$$\implies y' = Dy$$
where $y = S^{-1}x$. That last equation is a very easy system so solve; its equations are of the form $y_i' = \lambda_i y_i$, and so $y_i = c_i e^{\lambda_i t}$. Consequently, since
$$x = Sy$$
we have $x_i = S_i \cdot y$, by the definition of matrix multiplication, where $S_i$ is the $i^{th}$ row of $S$. This is equivalent to
$$x = v_1y_1+...+v_n y_n$$
as desired. 
A: Hint: Assume that we have destinct eigenvalues and eigenvectors. Write the solution as
$$x=v_1\exp(\lambda_1 t)+...+v_n\exp(\lambda_n t),$$
in which $\lambda_i$ is the $i$.th eigenvalue and $v_i$ is the i.th eigenvector.
Now, differentiate $x$ with respect to time.
$$\dot{x}=\lambda_1v_1\exp(\lambda_1 t)+...+\lambda_n v_n\exp(\lambda_n t)$$
Now, use $Av_i=\lambda_iv_i$ to eliminate the coefficients.
The rest should be obvious, after factoring $A$ and noting that the rest is equal to $x$.
