Set of solutions to first-order DE's is a subspace? A system of first-order linear homogeneous differential equations with constant coefficients is one of the form $$x'_1(t)=a_{11}x_1(t)+...+a_{1n}x_n(t)\\x'_2(t)=a_{21}x_1(t)+...+a_{2n}x_n(t)\\ ...\\x'_n(t)=a_{n1}x_1(t)+...+a_{nn}x_n(t)\\ $$ where $a_{ij}$ in R. A solution to this system is a collection of n functions $x_1(t),...,x_n(t)$ which can be written in vector form as $\textbf{x}(t)=\begin{bmatrix}x_1(t)\\...\\x_n(t)\end{bmatrix} $ so that the system of differential equations above can be written as $\textbf{x'}(t)=A\textbf{x}(t)$ where $A=[a_{ij}]$.
a) Show that the set of solutions of the system above is a subspace of $C^1(R)$, the space of continuously differentiable functions.
b) For a single equation of the form $f′(x) = kf(x)$, we know that $f(x) = ce^{kx}$ is a solution so we might make a guess that a solution to the system of equations above takes on the form $\textbf{x}(t) = e^{λt}\textbf{u}$. Show that $\textbf{x}$ is a solution if λ is an eigenvalue of A with corresponding eigenvector $\textbf{u}$.
Right so for part (a) I know what how you show something is a subspace but I'm confused as to what I'm looking at here. For instance, do you have to show that $\textbf{x}(t)=\textbf{0}$ is part of the set? And I'm also confused as to what part (b) is asking. Any help much appreciated!!
 A: Let $\mathbf y(t)$ and $\mathbf z(t)$ be solutions to
$\dot {\mathbf x}= A \mathbf x; \tag{1}$
that is,
$\dot {\mathbf y}(t) = A\mathbf y(t) \tag{2}$
and 
$\dot {\mathbf z}(t) = A\mathbf z(t); \tag{3}$
assuming $\mathbf y(t), \mathbf z(t) \in C^1(\Bbb R)$, for any $\alpha, \beta \in \Bbb R$ we have
$(\alpha {\mathbf y}(t) + \beta {\mathbf z}(t))' = \dfrac{d(\alpha {\mathbf y}(t) + \beta {\mathbf z}(t))}{dt} = \alpha {\mathbf y}'(t) + \beta {\mathbf z}'(t)$
$= \alpha A{\mathbf y}(t)  + \beta A{\mathbf z}(t) = A(\alpha {\mathbf y}(t) + \beta {\mathbf z}(t)); \tag{4}$
this shows $\alpha {\mathbf y}(t) + \beta {\mathbf z}(t)$ is a solution and hence the set of solutions form a linear subspace of $C^1(\Bbb R)$.  Note the solution $\mathbf w(t) = 0$ is covered here by taking $\alpha = \beta = 0$.
The assumption that in fact $\mathbf y(t), \mathbf z(t) \in C^1(\Bbb R)$ is validated by the standard existence and uniqueness theorems for ordinary differential equations, since it is easily seen that $A\mathbf x$ is globally Lipschitz continuous in $\mathbf x$:  $\Vert A\mathbf x_1 - A\mathbf x_2 \Vert = \Vert A(\mathbf x_1 - \mathbf x_2)\Vert \le \Vert A \Vert \Vert \mathbf x_1 - \mathbf x_2 \Vert $. See for example this wikipedia entry.
If
$A \mathbf u = \lambda \mathbf u, \tag{5}$
and we set
$\mathbf x(t) = e^{\lambda t} \mathbf u, \tag{6}$
then
$\dot {\mathbf x}(t) = \lambda e^{\lambda t} \mathbf u = e^{\lambda t} \lambda \mathbf u = e^{\lambda t}A \mathbf u = A( e^{\lambda t} \mathbf u), \tag{7}$
which shows (6) is a solution to (1).
