Is this a subspace with the given eigenvectors as its basis vectors? Given the eigenvectors:
$$\begin{pmatrix} 2 \\ 3 \end{pmatrix}, \begin{pmatrix}4 \\ 5\end{pmatrix}$$
And this linear combination equation
$$ \begin{pmatrix} 2 \\ 4 \end{pmatrix} = c_1 \begin{pmatrix} 2 \\ 3 \end{pmatrix}+c_2\begin{pmatrix}4 \\ 5\end{pmatrix}$$
Solving the equation gives: $$ c_1 = 3, c_2 = -1 $$
Which as far as I know could be written as the vector:
$$[v]_u = \begin{pmatrix}3 \\ -1 \end{pmatrix}$$
My question is, in this transition from the vector: \begin{pmatrix}2 \\ 4 \end{pmatrix} to this vector \begin{pmatrix}3 \\ -1 \end{pmatrix}
Is the latter vector $(3, -1)$ in a subspace consisting of the mentioned eigenvectors in the linear combination as the basis vectors for this subspace? I can't help but notice how similar it is to this linear combination in an euclidian space with it's basis vectors:
$$\begin{pmatrix} 2 \\ 4 \end{pmatrix} = 2 \begin{pmatrix}1 \\ 0 \end{pmatrix}+4 \begin{pmatrix}0 \\ 1 \end{pmatrix}$$
EDIT:
Let me rephrase the question to a more general one. If we have a linear combination $$ c_1 v_1+c_2v_2 $$
Could one say that this linear combination would create a subspace with the two vectors $v_1$ and $v_2$ as it's basis vectors? 

Note that $\frac{1}{2} \left[ 3 \begin{pmatrix} 4 \\ 5 \end{pmatrix} - 5 \begin{pmatrix} 2 \\ 3 \end{pmatrix} \right] = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $2 \begin{pmatrix} 2 \\ 3 \end{pmatrix} - \begin{pmatrix} 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, so the space with basis vectors $\Bigg \lbrace \begin{pmatrix} 4 \\ 5 \end{pmatrix} , \begin{pmatrix} 2 \\ 3 \end{pmatrix} \Bigg\rbrace$ is in fact $\mathbb{R}^2$. Hence every vector will be in the subspace with basis vectors the eigenvectors.
In general, $n$ linearly independent vectors will form a basis for $\mathbb{R}^n$, and in this case of $n = 2$ you do indeed have linearly independent vectors.
