I need to prove that the set of the solutions of the ODE $\dot{x} = Tx$ is a vector subspace of $C^1(\mathbb{R},\mathbb{R}^n)$ Let $T: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a linear transformation, $\mathbf{x}_0 \in \mathbb{R}^n$ and $t_0 \in \mathbb{R}$. Then the IVP
$$\begin{cases} 
\dot{\mathbf{x}} = T\mathbf{x} \\
\mathbf{x}(t_0) = \mathbf{x}_0
\end{cases}$$
has an unique solution in $\mathbb{R}$.
Using this theorem, i need to prove that the set of the solutions of the ODE $\dot{\mathbf{x}} = T\mathbf{x}$ is a vector subspace of the vector space $C^1(\mathbb{R},\mathbb{R}^n)$ with dimension = $n$.
I'm stucked in this problem since yesterday, and i still can't figure out how to show that.
Thanks in advance for any help.
Edit:
I figured out (with the helps from the answers below) how to show that the space of the solutions is a vector subspace. However i still have doubts on how to prove its dimension. I didnt understand very well why should we do the steps that Samuel presented. So here are my new questions:
1- Why should i build $n$ IVPs to build a basis to the subspace? And why the initial conditions should be the elements of the standard basis of $R^n$?
2- How to prove that the answers to the IVPs are LI each other? I guess that taking a determinant of the answers could work, right?
Thanks again for any help.
 A: Any solution to the ODE will be a continuous function from $\mathbb{R}$ to $\mathbb{R}^n$ (that is, an element of $C(\mathbb{R},\mathbb{R}^n))$, so in order to check that the set of solutions is a vector space you only have to show that the sum of solutions is a solution and the multiplication of a solution times scalar is a solution.
Let $x_1, x_2$ be solutions to the ODE, and let $\lambda\in\mathbb{R}$ be a scalar. Then

*

*If $z=x_1+x_2$, $$\dot{z}=\dot{x}_1+\dot{x}_2=Tx_1 + Tx_2 = T(x_1+x_2)=Tz$$
so $z$ is a solution to the ODE.

*If $w=\lambda x_1$,
$$\dot{w}=\lambda\dot{x}_1=\lambda Tx_1 =T(\lambda x_1)= Tw$$
so $w$ is a solution to the ODE.

Finally, since $\lambda$, $x_1$ and $x_2$ were arbitrary, we have proven that the set of solutions to the ODE is indeed a vector subspace of $C(\mathbb{R},\mathbb{R}^n)$.
Now, to see that it has dimension $n$, let's construct a basis. Let $x_1,...,x_n$ be the only solutions that satisfy $x_i(t_0)=\operatorname{e}_i$ (the $i$-th vector of the canonical basis of $\mathbb{R}^n$) for $i=1,\ldots,n$. As functions, they are linearly independent since their initial values are linearly independent. To conclude the proof, let's see that any solution to the ODE can be written as a linear combination of these $x_i$'s.
Let $y$ be a solution to the ODE, and let $v=(v_1,\ldots,v_n)=y(t_0)$. Now, let $z=\sum_{i=1}^n v_ix_i$. Note that $z(t_0)=\sum_{i=1}^n v_ix_i(t_0)=\sum_{i=1}^n v_i\operatorname{e}_i=y(t_0)$, so, by uniqueness of solutions, $y=z=\sum_{i=1}^n v_ix_i$. Hence $\{x_1,\ldots,x_n\}$ is a basis of the space of solutions to the ODE and therefore it has dimension $n$.
