How is it possible for a vector to be expressed with the same expression as its projection into another vector?

Question

I am a little confused with the following theorems (and resulting expressions):

1. If S={v₁, v₂, ...,v_n} 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, v_n)v_n

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

proj_Wu = (u, v₁)v₁ + (u, v₂)v₂ + ... + (u, v_n)v_n

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

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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.

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*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, v_n) = 0, allowing us to rewrite every (w₁, v_i) in the expression as (w₁ + w₂, v_i) = (u, v_i), for i=1, 2, ..., n.

Answer 1

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.