# 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?

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 $v \in V$.

• @ Umberto Please correct me if I'm wrong but a linear map on V should be defined for all elements of V. But does the extension in the answer not limit to vectors in V with $v_{m+1}=.....=v_n= 0$ thereby excluding vectors in V where these are not zero Sep 12 '13 at 18:31
• Since $\{w_1,\ldots,w_m,v_{m+1},\ldots,v_n\}$ is a basis of $V$, every vector $v \in V$ has a unique representation $v = \alpha_1 w_1 + \cdots + \alpha_m w_m + \alpha_{m+1}v_{m+1} + \cdots \alpha_n v_n$. Sep 13 '13 at 13:06
• Is the extension unique? Or there are other possible extensions? Aug 9 '16 at 20:29
• Say you have a map on the vector space of the x-axis that multiplies everything by 2. You can extend this to the whole plane to a map that multiplies everything by 2, or a map that multiplies only the x-axis by 2, etc. Nov 3 '19 at 5:05

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 $$V$$).