Theorem

This is one of the most important results in the whole book!

Suppose V is finite-dimensional and ${latex.inlineT \in L(V, W)}. Then range T is finite dimensional and ${latex.inline\text{dim V} = \text{dim null T} + \text{dim range T}}.

Proof

Let ${latex.inlineu_{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.inlineu_{1}, ..., u_{m}, v_{1}, ..., v_{n}} of V per 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.inlinev \in V}. Because ${latex.inlineu_{1}, ..., u_{m}, v_{1}, ..., v_{n}} spans V, we can write that ${latex.inlinev = 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.inlineTv_{1}, ..., Tv_{n}} is linearly independent. Suppose ${latex.inlinec_{1}, ..., c_{n}} are such that ${latex.inlinec_{1}v_{1}, + ... + c_{n}v_{n} = 0}. Then we get that ${latex.inlineT(c_{1}v_{1} + ... + T(c_{n}v_{n}) = T(c_{1}v_{1} + ... + c_{n}v_{n}) = 0} which implies that ${latex.inlinec_{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.