Linear Map extension from subspace to vector space $V$ is a finite dimensional vector space. How to prove that any linear map on subspace of $V$ can be extended to linear map on $V$.
I attempted it by taking the basis of the subspace $W$ with $\text{Dim}(m)$ and extending it to basis of $V$ with $\text{Dim}(n)$. Since mapping of all vectors of basis of $W$ exists and taking the remaining $m-n$ vectors as zero - we will get a linear map to an element that is in $V$. But is it sufficient to prove that all vectors in $V$ will also be having a linear map?
 A: If $\{w_1,\ldots,w_m\}$ is a basis of $W$, all you need to be able to do is show that there exist vectors $v_{m+1},\ldots,v_n \in V$ with the property that $\{w_1,\ldots,w_m,v_{m+1},\ldots,v_n\}$ is a basis of $V$. This is a pretty standard result in linear algebra.
The extension you propose $$\tilde L (v) = \tilde L (\alpha_1 w_1 + \cdots + \alpha_m w_m + \alpha_{m+1}v_{m+1} + \cdots \alpha_n v_n) = L(\alpha_1 w_1 + \cdots + \alpha_m w_m)$$ 
is linear and uniquely defined for all $v \in V$. 
A: Suppose $L:W\to W$ is the linear map that you want to extend to whole of $V$. First consider a basis $\{w_1, \ldots, w_m\}$ of $W$ and extend this to a basis $\{w_1, \ldots, w_m, v_{m+1}, \ldots, v_n\}$ of $V$. Now define $\widetilde{L}:V\to V $ as follows: we define the action of $\widetilde{L}$ on basis vectors by,
$$\widetilde{L}(w_1)= L(w_1),$$
$$\widetilde{L}(w_2)= L(w_2),$$
$$\vdots$$
$$\widetilde{L}(v_{m+1})=0,$$
$$\vdots$$
$$\widetilde{L}(v_n)= 0,$$
and on any arbitrary vectors $v\in V$, we define the action of $\widetilde{L}$ on $v=\alpha_1w_1+ \ldots + \alpha_mw_m + \alpha_{m+1}v_{m+1} + \ldots + \alpha_n v_n$, by
$$\widetilde{L}(v):= \alpha_1\widetilde{L}(w_1) + \ldots + \alpha_m\widetilde{L}(w_m) + \alpha_{m+1}\widetilde{L}(v_{m+1})+\ldots+ \alpha_n\widetilde{L}(v_n).$$
Its clear that $\widetilde{L}|_W = L$ and it is easy to check that $\widetilde{L}$ is a linear map (defined on whole of $V$).
