TITLE: Solving a system of ODEs using eignevalues
QUESTION [0 upvotes]: I want to make sure I understand the principles behind this and I know the best  way to know you have learned something is to be able to explain it to somebody else.  
Suppose I have a system $x'=Ax$. Where A is the coefficient matrix. The principle behind this is that if there there exists $\lambda$ such that  $Ax=\lambda x$then $x'=Ax=\lambda x$
Now this means that $Ax-\lambda x=0$ so $(A-\lambda I)x=0$ has a unique non-zero solution if an only if the determinant of $(A-\lambda I)$ is zero. Now my linear algebra is kind of weak, but is this akin to saying if we have two numbers $ab=0$, then either $a=0,b=0$, or both equal zero. 
Now setting the characteristic polynomial of $(A-\lambda I)$ to $0$ gives me eigenvalues and corresponding eigenvectors. And that combining these form the solutions for my original DE.  I am struggling to tie this information to the rest.

REPLY [1 votes]: So you start from a system of the form $x' = Ax$ with $x=\left( x_1(t) \,,\, x_2(t) \,,\, \ldots \,,\, x_n(t) \right)$.
Claim: if $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $x=ve^{\lambda t}$ is a solution of $x' = Ax$.
This is easy to check since:
$$x' = \lambda ve^{\lambda t} = Ave^{\lambda t} = Ax \quad\quad \square$$
The solution to a linear $n \times n$ system of differential equations is determined by $n$ linearly independent solutions; a linear combination gives you the general solution.
If $A$ has $n$ linearly independent eigenvectors $v_1 \,,\, v_2 \,,\, \ldots \,,\, v_n$ with corresponding eigenvalues $\lambda_1 \,,\, \lambda_2 \,,\, \ldots \,,\, \lambda_n$, then the general solution to $x'=Ax$ is given by:
$$x = \sum_{i=1}^n c_iv_ie^{\lambda_it}$$
This means that solving $x'=Ax$ comes down to finding the eigenvectors (and eigenvalues).
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