Linearly indepedent solutions to matrix differential equation x'=Ax True or false? 
If A is the 2-by-2 matrix 
\begin{bmatrix}0&1\\-a_0&-a_1\end{bmatrix}
and \begin{bmatrix}x_1,_1(t)\\x_1,_2(t)\end{bmatrix} and \begin{bmatrix}x_2,_1(t)\\x_2,_2(t)\end{bmatrix} are linearly independent solutions to the matrix  differential equation $x' = Ax$, then $x_{1,1(t)}$ and $x_{2,1 (t)}$ are linearly independent solutions to the differential equation  $y'' +a_1y' + a_0y = 0$, 
I tried matrix multiplication of each solution but I'm stuck at this point.
 A: I'm going to drop the "," in the subscripts and thus simplify expressions like $x_{1, 1}$ to $x_{11}$ und so weiter.
The above words being written:
Note that $(x_{11}(t), x_{12}(t))^T$ is a solution to 
$x' = Ax \tag{1}$
implies
$\begin{pmatrix}x_{11}(t)\\x_{12}(t)\end{pmatrix}' = \begin{bmatrix}0&1\\-a_0 &-a_1\end{bmatrix}\begin{pmatrix}x_{11}(t)\\x_{12}(t)\end{pmatrix}, \tag{2}$
or, writing (2) out in component form,
$x_{11}'(t) = x_{12}(t), \tag{3}$
$x_{12}'(t) = - a_0 x_{11}(t) - a_1 x_{12}(t); \tag{4}$
from (3),
$x_{11}''(t) = x_{12}'(t); \tag{5}$
now substituting (3) and (5) into (4) yields
$x_{11}''(t) = -a_0 x_{11}(t) - a_1 x_{11}'(t) \tag{6}$
or
$x_{11}''(t) + a_0 x_{11}(t) + a_1 x_{11}'(t), \tag{6}$
i.e., $x_{11}(t)$ satisfies
$y'' + a_1 y' + a_0 y = 0. \tag{7}$
Essentially the same computation shows that $x_{21}(t)$ also satisfies (7).
The linear independence of $x_{11}(t)$ and $x_{21}(t)$ is perhaps most easily seen by contradiction; for if there were nonzero $b, c \in \Bbb R$ with
$b x_{11}(t) + c x_{21}(t) = 0, \tag{8}$
then upon differentiating we find
$b x_{11}'(t) + c x_{21}'(t) = 0, \tag{9}$
and since by (3) and its corresponding equation for $x_{21}(t)$, $x_{21}'(t) = x_{22}(t)$, we find
$b x_{12}(t) + c x_{22}(t) = 0; \tag{10}$
(8) and (10) together may be written
$b \begin{pmatrix}x_{11}(t)\\x_{12}(t)\end{pmatrix} + c \begin{pmatrix}x_{21}(t)\\x_{22}(t)\end{pmatrix} = 0, \tag{11}$
in contradiction to the assumed hypothesis that $(x_{11}(t), x_{12}(t))^T$ and $(x_{21}(t), x_{22}(t))^T$ are linearly independent solutions to (1); thus (8) cannot bind and $x_{11}(t)$, $x_{21}(t)$ must be linearly independent solutions to (7).  
