This is one of the most important results in the whole book!
Suppose V is finite-dimensional and \({latex.inline[T \in L(V, W)](T \in L(V, W))}. Then range T is finite dimensional and \){latex.inline\text{dim V} = \text{dim null T} + \text{dim range T}}.
Let \({latex.inline[u_{1}, ..., u_{m}](u_{1}, ..., u_{m})} be a basis of null T, thus dim T = m. The linearly independent list \){latex.inlineu{1}, ..., u{m}} can be extended to a basis \({latex.inline[u_{1}, ..., u_{m}, v_{1}, ..., v_{n}](u_{1}, ..., u_{m}, v_{1}, ..., v_{n})} of V per [1753318250 - Axler 2.32 Every linearly independent list extends to a basis|2.32](1753318250 - Axler 2.32 Every linearly independent list extends to a basis|2.32). This is because the u’s are a basis of null T, null T is a subspace of V, and thus the u’s are a linearly independent list of vectors in V. Thus, \){latex.inline\text{dim} \ V = m + n}. We now want to show that range T is finitely dimensional and that its dimension is n. We do this by proving that ${latex.inlineTv{1}, ..., Tv{n}} is a basis of range T.
Let \({latex.inline[v \in V](v \in V)}. Because \){latex.inlineu{1}, ..., u{m}, v{1}, ..., v{n}} spans V, we can write that \({latex.inline[v = a_{1}u_{1} + ... + a_{m}u_{m} + b_{1}v_{1} + ... + b_{n}v_{n}](v = a_{1}u_{1} + ... + a_{m}u_{m} + b_{1}v_{1} + ... + b_{n}v_{n})}. We apply T on both sides to get that \){latex.inlineTv = a{1}Tu{1} + ... + a{m}Tu{m} + b{1}Tv{1} + ... + b{n}Tv{n} = b{1}Tv{1} + ... b{n}Tv{n}} because all of the u terms to go 0(since the u’s are in null T). In particular, this means the Tv’s span range T and thus range T is finite dimensional.
Now we want to show that \({latex.inline[Tv_{1}, ..., Tv_{n}](Tv_{1}, ..., Tv_{n})} is linearly independent. Suppose \){latex.inlinec{1}, ..., c{n}} are such that \({latex.inline[c_{1}v_{1}, + ... + c_{n}v_{n} = 0](c_{1}v_{1}, + ... + c_{n}v_{n} = 0)}. Then we get that \){latex.inline[T(c{1}v{1} + ... + T(c{n}v{n}) = T(c{1}v{1} + ... + c{n}v{n}) = 0](T(c{1}v{1} + ... + T(c{n}v{n}) = T(c{1}v{1} + ... + c{n}v{n}) = 0)} which implies that \({latex.inline[c_{1}v_{1} + ... + c_{n}v_{n} \in \text{null T}](c_{1}v_{1} + ... + c_{n}v_{n} \in \text{null T})} which means it can be written as a linear combination of u’s: \){latex.inlinec{1}v{1} + ... + c{n}v{n} = d{1}u{1} + ... + d{m}u{m} \rightarrow c{1}v{1} + ... c{n}v{n} - d{1}u{1} - ... - d{m}u{m} = 0}. But we know that the list of u’s and v’s is linearly independent, so all c’s must be 0. But then ${latex.inlineTv{1}, ..., Tv{n}} is linearly independent and hence is basis of T as desired.