Theorem

Every subspace of a finite-dimensional vector space is finite-dimensional.

Proof

Suppose V is finite-dimensional and U is a subspace of V. We need to prove that U is finite-dimensional. In other words, if V is finite-dimensional, we want to show that no matter what subspace we pick, we can find a list of finite length that spans that subspace. We do this through the following multi-step construction.

Step 1: If U = {0}, then U is finite dimensional and we are done. If \({latex.inline[U \ne 0](U \ne 0)} then choose a nonzero vector \){latex.inlineu_{1} \in U}.

Step k: If \({latex.inline[U = span(u_{1}, ..., u_{k-1})](U = span(u_{1}, ..., u_{k-1}))} then U is finite dimensional and we are done. If not, then choose a vector \){latex.inlineu{k} \notin span(u{1}, ..., u_{k-1})} and add it to the spanning list we are trying to find.

After each step, as long as the process continues, we have constructed a list of vectors such that no vector in this list is in the span of prevoius vectors. Thus after each step we have constructed a linearly independent list, by the linear dependence lemma. This linearly independent list cannot be longer than any spanning list of V. Thus the process eventually terminates, whcih means U is finite-dimensional.