Suppose V is finite-dimensional and U is a subspace of V. Then there is a subspace W of V such that ${latex.inlineV = U \bigoplus V}.
Since V is finite dimensional, so is U per 1753232272 - Axler 2.25 Finite dimensional subspaces|2.25. Thus there is a basis on U \({latex.inline[u_{1}, ..., u_{m}](u_{1}, ..., u_{m})} per [1753318233 - Axler 2.31 Every finite-dimensional vector space has a basis|2.31](1753318233 - Axler 2.31 Every finite-dimensional vector space has a basis|2.31). Obviously this basis is a linearly independent set of vectors in V. Thus it can be extended into a basis of V per [1753318250 - Axler 2.32 Every linearly independent list extends to a basis|2.32](1753318250 - Axler 2.32 Every linearly independent list extends to a basis|2.32) that looks like \){latex.inlineu{1}, ..., u{m}, w{1}, ..., w{n}}. Let \({latex.inline[W = span(w_{1}, ..., w_{n})](W = span(w_{1}, ..., w_{n}))}. Per [1753142442 - Axler 1.46 Direct sum of two subspaces|1.46](1753142442 - Axler 1.46 Direct sum of two subspaces|1.46), in order to show that \){latex.inlineV = U \bigoplus W} , we have to show that \({latex.inline[V = U + W](V = U + W)} and \){latex.inlineU \cap W = {0}} .
First Subproof Suppose \({latex.inline[v \in V](v \in V)}. Then because the list \){latex.inlineu+{1}, ..., u{m}, w{1}, ..., w_{n}} spans V, \({latex.inline[\exists a_{1}, ..., a_{m}, b_{1}, ..., b{n} \in F](\exists a_{1}, ..., a_{m}, b_{1}, ..., b{n} \in F)} such that \){latex.inlinev = a{1}u{1} + ... + a{m}u{m} + b{1}w{1} + ... + b{n}w{n}}. But the u’s are a spanning list of U, and the w’s are a spanning list of W, so \({latex.inline[a_{1}u_{1} + ... + a_{m}u_{m} \in U](a_{1}u_{1} + ... + a_{m}u_{m} \in U)} and \){latex.inlineb{1}w{1} + ... + b{n}w{n} \in W}. That proves the first result.
Second Subproof Now we show that \({latex.inline[U \cap W = \{0\}](U \cap W = \{0\})}. Suppose we have \){latex.inlinev \in U \cap W}. Then there is a way to express v as both a linear combination of the basis vectors of U and also a linear combination of the basis vectors of W i.e \({latex.inline[v = a_{1}u_{1} + ... + a_{m}u_{m} = b_{1}w_{1} + ... + b_{n}w_{n}](v = a_{1}u_{1} + ... + a_{m}u_{m} = b_{1}w_{1} + ... + b_{n}w_{n})}. But then \){latex.inlinea{1}u{1} + ... + a{m}u{m} - b{1}w{1} + ... + b{n}w{n} = 0}, which, due to the linear independence of the two bases combined, implies that each a and b = 0. Thus, v = 0. And we have our proof.