TITLE: Prove vector space is the direct sum of subspace and its orthogonal complement
QUESTION [8 upvotes]: $V$ is finite-dimensional over $\Bbb{C}$ and the form $\langle \cdot , \cdot \rangle$ is Hermitian. $U$ is a subspace of $V$.
Show that $V = U \oplus U^\perp$

I've been able to show that $U \cap U^\perp = \{0\}$. I don't know how to approach the problem showing that every vector $v\in V$ can be written as $v = u + u'$, where $u, u'$ are in $U$ and $U^\perp$ respectively.

REPLY [2 votes]: Choose any orthonormal basis $E=\{e_1 ,\ldots, e_k\}$ in $U$.  Continue $E$ to a basis $E^+ = \{e_1 ,\ldots, e_k;\; v_{k+1},\ldots, v_n\} $ in $V$.
Apply Gram-Schmidt to $E^+$. The first $k$ vectors in $E^+$ already are orthogonormal, and so Gram-Schmidt alters noting in them. The remaining $n-k$ vectors may change, and we get the orthonormal basis $E^{++} = \{e_1 ,\ldots, e_k;\; e_{k+1},\ldots, e_n\} $ in $V$.
The vectors $e_{k+1},\ldots, e_n$ are orthogonal to $e_1 ,\ldots, e_k$, and so they all are in $U^\perp$.
Present any $v\in V$ as the linear combination $v = (a_1 e_1 +\cdots+ a_k e_k)\;+\; (a_{k+1}e_{k+1}+\cdots + a_n e_n) = u+u'$. Clearly, $u \in U$ and $u' \in U^\perp$.