Inverse image of a vector subspace I've a question on the inverse image in linear functions.
Let $f:V\rightarrow W$ be a linear function between two vector spaces.
I know that if I have to find the inverse image of a particular vector $\vec{w} \in W$ I can find it as 
$f^{-1}(\vec{w})= \Big\{ { \vec{x_0}+\vec{y}}  \Big\}$ 
where $ \vec{x_0}$ is a particular vector contained in $f^{-1}(\vec{w})$ and $\vec{y}$ is a generic vector of $Ker(f)$, therefore $f^{-1}(\vec{w})$ is never a vector space.
But if I have to determine the inverse image of a vector subspace $\mathscr{K} \subset W$, where $\mathscr{K} =\mathscr{L}(\vec{v_1},\vec{v_2},...,\vec{v_k})$, can I find it as
$f^{-1}(\mathscr{K})=\mathscr{L} (\vec{y},\vec{x_1},\vec{x_2},...,\vec{x_k})$ 
where $\vec{y}$ is a generic vector of $Ker(f)$ and
$\vec{ x_1}$ is a vector such that $f(\vec{ x_1})=\lambda_1 \vec{v_1}$ ,
$\vec{ x_2}$ is a vector such that $f(\vec{ x_2})=\lambda_2\vec{v_2}$ and 
$\vec{ x_k}$ is a vector such that $f(\vec{ x_k})=\lambda_k \vec{v_k}$ ? 
In this way $f^{-1}(\mathscr{K})$ is a vector space as it should be.
Thanks a lot for your help

your guess about the inverse image of the subspace is right.
Take $\vec{x} \in f^{-1}(\mathscr{K})$, so that $f(\vec{x}) \in \mathscr{K}$.
Hence we will have,
\begin{align}
f(\vec{x})&= \sum \limits_{i=1}^{k}c_i \vec{v_i}\;\;\;\text{for some scalar} c_i\\
&= \sum \limits_{i=1}^{k}c_if(\vec{x_i})\\
&= f(\sum \limits_{i=1}^{k}c_i \vec{x_i})
\end{align}
This gives $\vec{x}-\sum \limits_{i=1}^{k}c_i \vec{x_i} \in kerf.$ And you get that $\vec{x}\in \mathscr{L}\{\vec{y},\vec{x_1},\vec{x_2},...,\vec{x_k}\}$.
