8. Rank-Nullity Theorem
(a) Let V be an n-dimensional vector space, and let W be an m-dimensional subspace of
V, where, of course, $m \le n$. Suppose that $B = \{b_1, \dots, b_m\}$ is a basis for W. Show
that V has a basis of the form $B \cup C$ for some set C of vectors. Hint: show that if A is
any linearly independent set of vectors, and c is vector not in the subspace spanned by
A, then $A \cup \{c\}$ is a linearly independent set. You may assume the property that any
collection of n linearly independent vectors from V is a basis for V.
(b) Let V and W be vector spaces defined over the same field F, and let $T: V \to W$ be
a linear transformation. Let B be a basis for the kernel, ker(T), of T and let $B \cup C$
be a basis for V. Let U be the subspace of V spanned by C. Show that U is mapped
injectively by T, i.e., for all $u, v \in U$ with $u \ne v$, we have $T(u) \ne T(v)$.
(c) Explain why the image of V under mapping by T is equal to the image of U under
mapping by T.
(d) Use the result of parts (b) and (c) to prove the rank-nullity theorem, which states that
$\text{dim(ker}(T)) + \text{dim(im}(T)) = \text{dim}(V)$, where im(T) denotes the image of V under mapping
by T.
