Exercise 3.A.11

Question

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 $S \in \mathcal{L}(U,W)$, then there exists $T \in \mathcal{L}(V,W)$ such that $Tu = Su$ for all $u \in U$.

Arden Shi · Accepted Answer

Because $U$ is a subspace of $V$, we can find $W$ such that
$$
V=U \bigoplus W
$$
So, by the definition of a direct sum, for any $v \in V$, $\exists u \in U, w \in W$ such that
\begin{equation}
v = u + w
\end{equation}
Now, define 
\begin{equation}
Tv = Su + w
\end{equation}
Now we need to prove that $T$ indeed is a linear map, and that $T$ extends $S$.
Let $v_1,v_2 \in V, u_1,u_2 \in U, w_1, w_2 \in W,v_1=u_1 + w_1, \uplambda \in F$ and $v_2=v_2+w_2$
To demonstrate additivity:
\begin{equation}
\begin{split}
\begin{align*}
T(v_1+v_2) &= S(u_1+u_2) + w_1+w_2 \
&= S(u_1) + w_1 + S(u_2) + w_2 \
&= T(v_1) + T(v_2)
\end{align*}
\end{split}
\end{equation}
To demonstrate homogeneity,
\begin{equation}
\begin{split}
\begin{align*}
T(\lambda v_1) &= S(\lambda u_1) + \lambda w_1 \
&= \lambda(S(u_1) + w_1) \
&= \lambda T(v_1) 
\end{align*}
\end{split}
\end{equation}
To prove that $T$ extends $S$,

By (1)
$$
v = u + w
$$
Therefore, when $v \in U$, $u = v$ satisfies the equation. By the definition of a direct sum, we can conclude that $u = v$ is the only possible solution, therefore $w = 0$.  Thus, when substituting $w = 0$ into equation (2), we obtain 
$$
Tu = Su
$$
\qed

Exercise 3.A.11

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Exercise 3.A.11

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