How do I solve systems such as $\phi' = A \phi$, where $\phi$ and $A$ are matrices? So I have to solve 
$\phi' = A \phi, \phi(0) = \mathbb{I}$
where
$A = \begin{bmatrix}
0 && 1\\
-a^2 && 0
\end{bmatrix}$
How do I begin to solve this?
 A: The solution is $\phi = e^{At}$ (yes, it works just as well for matrices as for real functions).
If you haven't encountered it, the definition of $e^B$ for a matrix $B$ is
$$
\Bbb I + B + \frac{B^2}2 + \frac{B^3}6 + \cdots
$$
just as for real numbers. If you calculate the first few terms in the power series explicitly you should see a pattern indicating that the elements of $\phi$ are sines and cosines.
A: @Arthur After a remark of @egreg here is a corrected explicit expression for $exp(tA)$ (with the aid of Matlab)
$$e^{tA}=\begin{bmatrix}\cos(ta)&\dfrac{1}{a}\sin(ta)\\-a \sin(ta)&\cos(ta)\end{bmatrix}$$
involving circular functions...only :)
A: The solution is indeed
$$
\phi(t)=e^{tA}=I+tA+\frac{t^2}{2!}A^2+\frac{t^3}{3!}A^3+\dotsb
$$
In order to compute the powers of $A$ you can diagonalize the matrix; its eigenvalues are the roots of $X^2+a^2$ so they are $ia$ and $-ia$.
The diagonalizing matrix is
$$
\begin{bmatrix}
1 & 1 \\
ia & -ia
\end{bmatrix}
$$
whose inverse is
$$
-\frac{1}{2ia}
\begin{bmatrix}
-ia & -1 \\
-ia & 1
\end{bmatrix}
$$
so we have
\begin{align}
A^n
&=-\frac{1}{2ia}
\begin{bmatrix}
1 & 1 \\
ia & -ia
\end{bmatrix}
\begin{bmatrix}
(ia)^n & 0 \\
0 & (-ia)^n
\end{bmatrix}
\begin{bmatrix}
-ia & -1 \\
-ia & 1
\end{bmatrix}
\\[6px]
&=
\frac{1}{2ia}
\begin{bmatrix}
(ia)^{n+1}-(-ia)^{n+1} & (ia)^n-(-ia)^n \\
(ia)^{n+2}-(-ia)^{n+2} & (ia)^{n+1}-(-ia)^{n+1} \\
\end{bmatrix}
\end{align}
The series for position $(1,1)$ is
$$
\sum_{n\ge0}\frac{(iat)^n+(-iat)^n}{2n!}=\frac{e^{iat}+e^{-iat}}{2}=\cos(at)
$$
The series for position $(1,2)$ is
$$
\frac{1}{ia}\sum_{n\ge0}\frac{(iat)^n-(-iat)^n}{2n!}
=\frac{e^{iat}-e^{-iat}}{2ia}=\frac{1}{a}\sin(at)
$$
At position $(2,1)$ we have the same as in position $(1,2)$, but multiplied by $-a^2$, so $-a\sin(at)$. Thus
$$
e^{At}=
\begin{bmatrix}
\cos(at) & \frac{1}{a}\sin(at) \\
-a\sin(at) & \cos(at)
\end{bmatrix}
$$
A: After a night, a rather elementary fact that I had completely overlooked yesterday, and might be of some interest for you, HMParticle, and maybe for those having worked on the issue (@egreg, @Arthur ...) . In fact, the (first order) matrix-vector system $\Phi' = A \Phi$, written under the form:
$$\begin{cases}x'&=& & y\\y'&=&-a^2x& \end{cases} \ \ (1)$$
is equivalent to the classical harmonic oscillator differential (second order) equation, with appropriate initial conditions, i.e.,
$$y''=-a^2y \ \ (2)$$
The proof is straightforward: derive the second equation of system (1): $y''=-a^2x'$, then plug in it the expression of $x'$ coming from the first equation to obtain relationship (2).
