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The <orthogonal complement> is $W^\perp=\{v\in V:\langle v,w\rangle=0\text{ for every }w\in W\}$. Put $Pv=\sum_i\langle v,e_i\rangle e_i$, so $T=I-P$. <Orthonormality> gives $P^2=P$, and therefore $T^2=T$. Also $\langle Tv,e_j\rangle=0$ for every $j$, so $Tv\in W^\perp$. Every $v=Pv+Tv$ is thus a sum from $W$ and $W^\perp$, and their intersection is zero, since a vector <orthogonal> to itself has squared <norm> zero. This proves $V=W\oplus W^\perp$. The decomposition is unique, so its two <orthogonal projection>[projection] maps depend only on $W$, not on the chosen <orthonormal basis>.
Finite dimensionality of $W$ cannot simply be dropped. Take the real <Hilbert space> $V=\ell^2(\mathbb N;\mathbb R)$ and $W=c_{00}$, the finitely supported sequences. This is an infinite-dimensional proper <vector subspace>[subspace]. A vector <orthogonal> to every coordinate <unit vector> is zero, so $W^\perp=\{0\}$. Yet, for example, $(2^{-j})_{j\ge1}$ belongs to $V\setminus W$. Hence $W\oplus W^\perp=W\ne V$. The failure occurs even though $V$ itself is complete; the proper <dense subspace> $W$ is not closed.
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