Extending linear maps from subspaces to the entire space

Question

This is from Axler's Linear Algebra Done Right, 3.A.11

Suppose $V$ is finite-dimensional. Prove that every linear map on a subspace of $V$ can be extended to a linear map on $V$. In other words, show that if $U$ is a subspace of $V$ and $S \in L( U, W$), then there exists $T \in L (V,W)$ such that $T(u) = S(u) $ for all $u \in U$.

I tried to prove this by creating another transformation that took all vectors into U and then mapped all vectors in $U$ into $S$($U$). (So a double transformation I guess.) The answer book had a different solution which I didn't understand:

I don't see how this is linear, because consider $u + v$, where $u \in U$, $S$($u$) $\ne 0 $ and $v \in V$ and $u + v \in V \notin U $. (since $U \subset V$).

Then $T$($u+v$) $=0$, but $T$($u+v$) $=0$, but $T$($u$) $+ T$($v$) $=$ $S$($u$) $\ne 0 $.

Which means that T is not linear.

Answer 1

This is really a matter of applying the theorem you asked about in your previous question. Your previous theorem said (I modify the notation slightly)

Let $V$, W be vector spaces over a field $F$, and let $\{\xi_1, \dots, \xi_n\}$ be a basis of $V$. Also, choose $n$ vectors $w_1, \dots, w_n \in W$. Then, there is a unique linear transformation $T:V \to W$ such that for every $i \in \{1, \dots, n\}$, we have $T(\xi_i)= w_i $

So, now you started with a basis $\{u_1, \dots, u_m\}$ of $U$, and then you extend it to a basis $\{u_1, \dots, u_m, v_{m+1}, v_n\}$ of $V$. Now, to make it super explicit how to directly apply the theorem stated above, let's rename things: define \begin{equation} \xi_i = \begin{cases} u_i& \text{if $1 \leq i \leq m$} \\ v_i & \text{if $m+1 \leq i \leq n$} \end{cases} \end{equation} and define \begin{equation} w_i = \begin{cases} S(u_i)& \text{if $1 \leq i \leq m$} \\ 0 & \text{if $m+1 \leq i \leq n$} \end{cases} \end{equation}

Then, by the theorem, there exists a (unique) linear map $T: V \to W$, such that for all $i \in \{1, \dots, n\}$, $T(\xi_i) = w_i$.

Now, to show $T$ restricted to $U$ equals $S$, i.e for all $x \in U$, $T(x) = S(x)$, notice that BY DEFINITION, for all $i \in \{1, \dots, m\}$ we have that $T(u_i) = S(u_i)$. Since $T$ and $S$ agree on the basis $\{u_1, \dots u_m\}$ of $U$, they agree everywhere on $U$ (this is the uniqueness argument I presented in the previous answer).

Answer 2

I think it is also good to get an intuition for the solution. The easiest way to think about this is to make T a projection of V onto U (think about it in 3D space: if U is the xy plane, just "flatten" everything onto the plane). So, if you have a basis of U and extend it to a basis of V, you make T just "take the scalars that multiply the vectors in the basis of U and discard the rest". This "flattens" (whatever that means in spaces other than R3) everything onto U. And this is exactly what the solution from the answer book does, with all the algebra written down.