TITLE: Solving a matrix as a differential equation
QUESTION [6 upvotes]: If I have a matrix
$$A = \begin{bmatrix}5 & 4 & -6\\-2 & -1 & 2\\2 & 0 & -3\end{bmatrix}$$
how do I solve $x'=Ax$ as a differential equation? My text book explains this in a rather confusing way and I am really not getting it - what if I just found the eigen values - would the solution be anything to do with that or could I just use some kind of gaussian elimination?
Any help would be much appreciated, many thanks. :)

REPLY [4 votes]: The solution to your differential equation is $$x(t) = e^{At}$$ where $e^{A}$ is defined as $$e^A = \sum_{i=0}^{\infty} \frac{A^k}{k!}.$$ This infinite series can be evaluated by writing $A = PDP^{-1}$, the eigenvalue decomposition of A (assuming it exists), where $D$ is a diagonal matrix of eigenvalues of $A$. For an example, see http://www.millersville.edu/~bikenaga/linear-algebra/matrix-exponential/matrix-exponential.html.

REPLY [3 votes]: You need the Eigenvalues, the main idea is that 
$$\frac{d}{dt} e^{At}= A e^{At}$$ 
To compute $e^{At}$ you need the eigenvalues and their multiplicity, gaussian elimenation won't help you.
The Eigenvalues of your matrix are 
$$\sigma=\{ 1+2i,1-2i,1\}$$ 
and the eigenvectors are 
$$\begin{pmatrix}  2+i \\ - 1 \\ 1 \\ \end{pmatrix} \quad \begin{pmatrix} 2-i\\ -1\\ 1 \end{pmatrix} \quad \begin{pmatrix} 1\\ 0 \\ 1 \end{pmatrix}$$
So the system is an linear combination of $e^{\lambda_i t} v_i$.