Second order differential equation by linear algebra I am trying to solve the following differential equations $\frac {du}{dt} = Au$ and $\frac {d^2u}{dt^2} = Au$ with 
$$
A = 
        \begin{bmatrix}
        -2 & 1 & 0 \\
        1 & -2 & 1 \\
        0 & 1 & -2 \\
        \end{bmatrix},
$$
$u(0) = (0, 2 \sqrt 2, 0)$ and $\frac{du}{dt}(0) = 0$
by linear algebra. I get the complete solution for the first equation and get stuck on the second one. The textbook ("Introduction to Linear Algebra" by Strang) offers the following solution:
$\frac {d^2u}{dt^2} = Au$ has the same eigenvectors  x.  But  now the eigenvalues  $\lambda$ lead to oscillations $e^{i\omega t}x$ and $e^{-i\omega t}x$.  The  frequencies  come  from $\omega^2 = -\lambda$.
Then it proceeds to plug $e^{i\omega t}x$ into the second order equation to show the last equation holds. My question is:
Is there a specific method for solving this kind of second order equation relying on $e^{i\omega t}x$? Or am I missing something and there actually is a way to deduce $e^{i\omega t}x$ from what is asked? Needles to say, at this point in the material, specific methods for solving the second equation have not been covered.
 A: We can solve the second order system by thinking of it as a larger first order system.  In particular, define $v = \frac{du}{dt}$.  We can rewrite your one matrix equation as a "system of matrix equations", namely
$$
\frac {du}{dt} = v\\
\frac{dv}{dt} = Au
$$
We can write the whole thing with one matrix using block matrices.  In particular, we have
$$
\frac{d}{dt} \pmatrix{u\\v} = \pmatrix{0&I\\A&0}
$$
That is, if we let $x$ denote the column-vector $(u_1,u_2,u_3,v_1,v_2,v_3)$, then our linear differential equation is $\frac {dx}{dt} = Mx$, where 
$$
M = \pmatrix{
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1\\
-2 & 1 & 0 &0&0&0\\
1 & -2 & 1 &0&0&0\\
0 & 1 & -2 &0&0&0\\
}
$$
with the initial condition $x(0) = (0,2\sqrt{2},0,0,0,0)$.  You should be able to take it from there on your own and reach the desired conclusion by rote.

If we want to figure out why those new eigenvalues should satisfy that equation, suppose that $\omega$ is an eigenvalue of $M$, and $y$ is an eigenvector.  Split $y$ up into the two vectors of length $3$ $y_1$ and $y_2$.  note that
$$
My = \omega y \implies\\
\pmatrix{0&I\\A&0}\pmatrix{y_1\\y_2} = \omega\pmatrix{y_1\\y_2} \implies\\
\pmatrix{y_2\\ Ay_1} = \pmatrix{\omega y_1\\ \omega y_2}
$$
With this system of equations, we should be able to deduce that $y_1$ satisfies $Ay_1 = \omega^2 y_1$.  That is, $\omega^2$ should be an eigenvalue of $A$.
