= Solution
The <rank-nullity theorem> says that for a linear map $T:V\to Z$ with finite-dimensional domain,
$$
\dim V=\dim\ker T+\dim\operatorname{im}T.
$$
The <direct sum> $U\oplus W$ consists of pairs $(u,w)$ with componentwise vector-space operations. The canonical inclusions are
$$
i(u)=(u,0),
\qquad
j(w)=(0,w).
$$
For subspaces $U,W\leq V$, define
$$
q:U\oplus W\longrightarrow U+W,
\qquad
q(u,w)=u+w.
$$
This is surjective, and
$$
\ker q=\{(x,-x):x\in U\cap W\}
=\operatorname{im}l.
$$
The <First isomorphism theorem for vector spaces> therefore gives
$$
\boxed{(U\oplus W)/\operatorname{im}l\cong U+W\leq V}.
$$
Back to article page
