Theorem

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}.

Proof

Since V is finite dimensional, so is U per 2.25. Thus there is a basis on U ${latex.inlineu_{1}, ..., u_{m}} per 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 2.32 that looks like ${latex.inlineu_{1}, ..., u_{m}, w_{1}, ..., w_{n}}. Let ${latex.inlineW = span(w_{1}, ..., w_{n})}. Per 1.46, in order to show that ${latex.inlineV = U \bigoplus W} , we have to show that ${latex.inlineV = U + W} and ${latex.inlineU \cap W = {0}} .

First Subproof Suppose ${latex.inlinev \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} 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.inlinea_{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.inlineU \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.inlinev = 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.