How is it possible for a vector to be expressed with the same expression as its projection into another vector? I am a little confused with the following theorems (and resulting expressions):


*

*If S={v1, v2, ...,vn} is an orthonormal basis for an inner product space V, and if u is any vector in V, then



u = (u, v1)v1 + (u, v2)v2 + ... + (u, vn)vn



*If W is a finite-dimensional subspace of an inner product space V, and {v1, v2, ...,vn} is an orthonormal basis for W, and u is any vector in V, then



projWu = (u, v1)v1 +
  (u, v2)v2 + ... + (u,
  vn)vn

(explanation of this dervition, if needed, is given below)*

What confuses me, is that both expressions are the same, and it seems to me that both the vector u, and its projection on w1 is the same. How does this make sense? How can a vector and its projection be expressed by the same expression?
I'd appreciate any help/comment/explanation regarding this. 

*The expression in 2 is dervied as follows: vector u is expressed as the sum of two orthogonal vectors w1 and w2. Vector w1 is then decomposed according to theorem 1 above, and because w2 is orthogonal to W, it follows that (w2, v1) = (w2, v2) = ... = (w2, vn) = 0, allowing us to rewrite every (w1, vi) in the expression as (w1 + w2, vi) = (u, vi), for i=1, 2, ..., n. 
 A: Let's say we have a finite dimensional inner product space $V$ and a subspace $W \subseteq V$ and let $\mathbf{u} \in V$. If we want to apply the first statement, we need some orthonormal basis $\mathbf{v}_1, \dots, \mathbf{v}_n$ of $V$ and then we get
$$ \mathbf{u} = \sum_{i=1}^n (\mathbf{u}, \mathbf{v}_i) \mathbf{v}_i. $$
If we want to apply the second statement, we need an orthonormal basis of $W$, not of $V$! Let us choose such a basis $\mathbf{w}_1,\dots,\mathbf{w}_m$ (with $m \leq n$). Then
$$ \operatorname{proj}_W(\mathbf{u}) = \sum_{i=1}^m (\mathbf{u}, \mathbf{w}_i) \mathbf{w}_i $$
so you can see that the expressions are different. Actually, you can always choose an orthonormal basis $\mathbf{v}_1,\dots,\mathbf{v}_n$ of $V$ such that the first $m$ vectors $\mathbf{v}_1,\dots,\mathbf{v}_m$ form an orthonormal basis of $W$ and then you get the equations
$$ \mathbf{u} = \sum_{i=1}^n (\mathbf{u}, \mathbf{v}_i) \mathbf{v}_i,\\
\operatorname{proj}_W(\mathbf{u}) = \sum_{i=1}^m (\mathbf{u}, \mathbf{v}_i) \mathbf{v}_i $$
but they are still different because the summation in the first goes up to $n$ while the summation in the second goes up to $m$. If $m = n$ then $V = W$ and indeed $\operatorname{proj}_W(\mathbf{u}) = \mathbf{u}$ but if not, you get different expressions.
