Suppose $\mathcal V\subset \mathbb{R}^n$ is a subspace and $\vec{x}\in\mathbb{R}^n$. I want to show there is a unique way to write $\vec{x}$ as a sum of two vectors $\vec{x}=\vec{x}^{\parallel}+\vec{x}^{\perp},$ the first of which lies in $\mathcal{V}$ and the second of which is orthogonal to every member of $\mathcal{V}.$
This is a simple problem but I'm just wondering if a step in the proof in my book is wrong.
The proof begins by letting $Y=\{\vec{y}_1...,\vec{y}_k\}$ be an orthonormal basis for any $k$-dimensional subspace $\mathcal V$ of $\mathbb{R}^n.$ We then must show that there exists a unique choice of scalars $a_1, ... , a_k\in\mathbb{R}$ such that if we define $\vec{x}^{\parallel}=a_1\vec{y}_1+...+a_k\vec{y}_k$ and $\vec{x}^{\perp}=\vec{x}-\vec{x}^{\parallel}$ then $\vec{x}^{\perp}$ is orthogonal to every vector in $\mathcal{V}.$ However, the next step of logic is what I don't understand.
Because the inner product is bilinear, that is the same as saying that $\vec{x}^{\perp}$ is orthogonal to every element of the basis $Y$ of $\mathcal{V}$.
To me, it seems that if $\vec{x}^{\perp}$ is orthogonal to every element of $\mathcal{V}$, then certainly it's orthogonal to the basis elements $Y$ since they are elements of $\mathcal{V}$. However, the book appeals to bilinearity. The only way I could see that being relevant if is we let $\vec{v}$ be an arbitrary element of $\mathcal{V}$ so that $\vec{v}=c_1\vec{y}_1+...c_k\vec{y}_k$ for some $c_1, ..., c_k\in\mathbb{R}$; then $$\langle\vec{x}^{\perp},\vec{v}\rangle=0\iff\langle\vec{x}^{\perp},c_1\vec{y}_1+...+c_k\vec{y}_k\rangle=c_1\langle\vec{x}^{\perp},\vec{y}_1\rangle+...+c_k\langle\vec{x}^{\perp},y_k\rangle=0$$ but the latter does not imply $\langle \vec{x}^{\perp}, \vec{y}_i\rangle=0$ for each $i$ up to $k$ as far as I can see.
So what did the author mean by appealing to bilinearity?
To be clear, this is in Chapter 1 Section 2 of Kristopher Tapp's Differential Geometry of Curves and Surfaces textbook.
Let us write $x \perp y$ for $\langle x, y \rangle = 0$. The important equivalence here is: $$u \perp \mathcal{V} \iff u \perp y_i \text{ for each } i = 1, \dots, k,$$ where the left side means that $u$ is perpendicular to every element of $\mathcal{V}$. Now since each $y_i \in \mathcal{V}$, the $\implies$ direction does not require anything specific about the definition of orthogonality. But you need bilinearity to go the other way (to be pedantic, only the linearity in the second variable): expand an arbitrary $v \in \mathcal{V}$ as $\sum c_i y_i$ and write $$\langle u, v \rangle = \sum c_i \langle u, y_i \rangle = 0.$$ As you note, if you know that $u \perp v$ for a specific $v$, ie a specific choice of $c_1, \dots, c_k$, that does not imply that $u \perp y_i$ for each $i$.
But the above equivalence means that to check a tentative decomposition $x = x^\parallel + x^\perp$ has the required properties, it suffices to check that $x^\perp \perp y_i$ for the basis elements $y_i$ (and $x^\parallel \in \mathcal{V}$, which is equivalen to saying that $x^\parallel$ is of the form $\sum a_i y_i$). Since $y_i$ are chosen to be orthonormal, you can now check that $a_i = \langle x, y_i \rangle$ is the unique choice of $a_i$ that makes this work, which therefore gives the unique such decomposition of $x$.