Proving any linear map on a subspace of $V$ can be extended to a linear map on $V$

Question

Suppose V is a finite dimensional vector space. I am trying to prove that any linear map on subspace of V can be extended to linear map on V.

Basically, showing that if $U$ is a subspace of $V$ and $S \in L(U,W)$, then there exists a $T \in L(V,W)$ such that $Tu=Su$ for all $u \in U$.

I attempted it by taking the basis of the subspace U with Dim(m) and extending it to basis of V with 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 by defining $T \in L(V,W)$ as:

$T(a_1u_1 + ...+ a_mu_m+b_1v_1 + ...+b_nv_n)=a_1Su_1 + ...+a_mSu_m$

Then $Tu=Su \space \forall u \in U$

What I dont get is this:

1) 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 inside $T$ not limit to vectors in V with $v_m+1=.....=v_n=0$ thereby excluding vectors in V where these are not zero?

2) How does the $Su$'s hit every value in $W$?

3) Is there a more intuitive way or step by step explained way of how I need the above equation and condition? I have no idea why it was chosen as thus.

Thanks!

Answer 1

I think there's a much easier way to go about this. Let $W$ be a proper subspace of $V$. Then $W$ has a basis $\{w_1,\dotsc,w_k\}$. A standard theorem of linear algebra says this basis for $W$ can be extended to a basis $\{w_1,\dotsc,w_k,v_1,\dotsc,v_n\}$ for $V$. Now, let $f:W\rightarrow U$ be a linear map. Then let $\overline{f}(w_i)=f(w_i)$ for $1\leq i\leq k$ and let $\overline{f}(v_i)=\mathbf0$ for $1\leq i\leq n$. Extending linearly gives an extension $\overline{f}:V\rightarrow U$ of $f$ to $V$.

Answer 2

The OP, I assume, knows this, but in the interest of making this question more searchable, this is Exercise 3.3 in Axler's Linear Algebra Done Right (as well as the reason I was looking up questions like this).

I can only imagine what grading twenty or thirty proofs for this (and for a few other results) does to one's mind, but apparently my professor didn't find anything wrong with mine (but for a few typographical errors, now corrected). I'm posting it because, from what I can tell, it differs somewhat from the solutions so far; specifically, it seems a little more general in that it doesn't assign a specific value for certain preimages. I'm still far from being comfortable with any of this stuff, though, so if I'm wrong on that point, please correct me. Anyway —

Because $U$ is a finite-dimensional subspace, it has a basis — call it $B=\left\{u_{1},u_{2},\ldots u_{r}\right\}$. Moreover, since $B$ is linearly independent in $V$, it can be expanded to a basis for $V$ — say, $B^{\prime}=\left\{ u_{1},\ldots u_{r},v_{r+1},\ldots v_{s}\right\}$. Since $B^{\prime}$ is also linearly independent, we can form a basis for another subspace, $V^{\prime}$, with the vectors in $B^{\prime}\backslash B=\left\{ v_{r+1},\ldots v_{s}\right\}$. Let $T^{\prime}\in\mathcal{L}\left(V^{\prime},W\right)$. Now we define $T:V\rightarrow W$ such that $T\left(v\right)=S\left(u\right)+T^{\prime}\left(v^{\prime}\right)$. Because $U$ and $\text{span}(B^{\prime}\backslash B)$ are subspaces of $V$, they both belong to $\mathcal{L}\left(V,W\right)$ as well; and since $\mathcal{L}\left(V,W\right)$ forms a vector space, $S+T^{\prime}=T$ is linear as well. Since each $u\in U$ can be written in $V$ as $a_{1}u_{i}+\ldots+a_{r}u_{r}+b_{r+1}v_{r+1}+\ldots+b_{s}v_{s}$, where $a_{i}\in\mathbb{F}$ and $b_{j}=0$, $T\left(u\right)=S\left(u\right)+T^{\prime}\left(0\right)=S\left(u\right)$.

Answer 3

I think that introducing the projection map is more intuitive. Given a vector space $V$, a linear map $P: V \to V$ is said to be a linear projection if $P^{2} = P$. As a consequence of this definition, given a subspace $U$ of $V$, there exists a linear projection $P:V \to V$ such that $P(V) = U$. To show this is not difficult, just consider $U'$ a linear complement of $U$, so $V = U\bigoplus U'$, and given $v \in V$ we can express in a unique way $v = x + y$ where $x\in U$, $y\in V$; finally define $P(v) = x$ and show that $P$ is linear and $P^{2} = P$.

To answer the question, just define $T:V \to W$ as $T = S\circ P$. Note that $T(u) = S(u)$ for all $u \in U$.