Rewriting inner product in terms of linear combinations of orthonormal basis vectors

Question

If I have an orthonormal basis $B = \{ \pmb {b_{1}}, \pmb {b_{2}}, \dots, \pmb {b_{n}} \}$ for a finite-dimensional inner product space $V$, then it says here that every $\pmb v \in V$ can be written as $\pmb v = \langle \pmb{b_{1}}, \pmb v \rangle \pmb{b_{1}} + \langle \pmb{b_{2}}, \pmb v \rangle \pmb{b_{2}} + \dots + \langle \pmb{b_{n}}, \pmb v \rangle \pmb{b_{n}}$, therefore the corresponding coordinate vector should be $$[\pmb v]_{B} = \bigg(\langle \pmb{b_{1}}, \pmb v \rangle, \langle \pmb{b_{2}}, \pmb v \rangle, \dots, \langle \pmb{b_{2}}, \pmb v \rangle \bigg)$$with each coordinate being $v_{i} = \langle \pmb{b_{i}}, \pmb v \rangle$. And so I decided to expand out $\langle \pmb u, \pmb v \rangle$ as: $$ \begin{align} \langle \pmb u, \pmb v \rangle &= \bigg\langle u_{1} \pmb{b_{1}} + u_{2} \pmb{b_{2}} + \dots + u_{n} \pmb{b_{n}}, \ \pmb v \bigg\rangle \\ \\ &= u_{1} \langle \pmb{b_{1}},\pmb{v} \rangle + u_{2} \langle \pmb{b_{2}}, \pmb{v} \rangle + \dots + u_{n} \langle \pmb{b_{n}}, \pmb{v} \rangle \\ &= \sum_{i = 1}^{n} u_{i} \langle \pmb{b_{i}}, \pmb v \rangle \\ \end{align} $$ Here, I am really tempted to just replace $\langle \pmb{b_{i}}, \pmb v \rangle$ with $v_{i}$, and if I do so I get this: $$ \begin{align} \langle \pmb u, \pmb v \rangle &= \sum_{i = 1}^{n} u_{i} \langle \pmb{b_{i}}, \pmb v \rangle \\ &= \sum_{i = 1}^{n} u_{i} v_{i} \\ &= [\pmb u]_{B} \cdot [\pmb v]_{B} \end{align} $$ And so the inner product of two vectors in $V$ would be the dot product of the corresponding coordinate vectors over the basis $B$.

But, if I don't substitute, and instead I keep expanding, I get this: $$ \begin{align} \langle \pmb u, \pmb v \rangle &= \sum_{i = 1}^{n} u_{i} \langle \pmb{b_{i}}, \pmb v \rangle \\ &= \sum_{i = 1}^{n} u_{i} \bigg\langle \pmb{b_{i}}, \ v_{1} \pmb{b_{1}} + v_{2} \pmb{b_{2}} + \dots + v_{n} \pmb{b_{n}} \bigg\rangle \\ &= \sum_{i = 1}^{n} u_{i} \bigg( \overline{v_{1}} \langle \pmb{b_{i}}, \pmb{b_{1}} \rangle + \overline{v_{2}} \langle \pmb{b_{i}}, \pmb{b_{2}} \rangle + \dots + \overline{v_{n}} \langle \pmb{b_{i}}, \pmb{b_{n}} \rangle \bigg) \\ &= \sum_{i = 1}^{n} u_{i} \bigg( \sum_{j = 1}^{n} \overline{v_{j}} \langle \pmb{b_{i}}, \pmb{b_{j}} \rangle \bigg) = \sum_{i = 1}^{n} \sum_{j = 1}^{n} u_{i} \overline{v_{j}} \langle \pmb{b_{i}}, \pmb{b_{j}} \rangle \end{align} $$ We know that $\langle \pmb{b_{i}}, \pmb{b_{i}} \rangle = 1$, and if $i \ne j$ then $\langle \pmb{b_{i}}, \pmb{b_{j}} \rangle = 0$. Therefore non-zero terms occur only when $i=j$, so: $$ \begin{align} \langle \pmb u, \pmb v \rangle &= \sum_{i = 1}^{n} \sum_{j = 1}^{n} u_{i} \overline{v_{j}} \langle \pmb{b_{i}}, \pmb{b_{j}} \rangle = \sum_{i = 1}^{n} u_{i} \overline{v_{i}} \langle \pmb{b_{i}}, \pmb{b_{i}} \rangle \\ &= \sum_{i = 1}^{n} u_{i} \overline{v_{i}} \end{align} $$ And so this way, I get a different expression for $\langle u, v \rangle$, which could potentially mean:

1. $v_{i} = \overline{v_{i}}$, which implies $\langle \pmb{b_{i}}, \pmb v \rangle = \langle \pmb v, \pmb{b_{i}} \rangle$.
2. The first method is wrong. Perhaps there's a subtlety where you can't "just replace" $\langle \pmb{b_{i}}, \pmb v \rangle$ with $v_{i}$.
3. The second method is wrong. Perhaps I made a mistake somewhere, when expanding.
4. Both methods are wrong.

Is it any of these? Whats going on.

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EDIT: as per @RyeCatcher's comments, my decomposition of $\pmb v$ leads to $\langle \pmb{b_{i}}, \pmb v \rangle = \langle \pmb v, \pmb{b_{i}} \rangle$, which is just wrong; the correct formula should include complex conjugation.

From this I am guessing that the correct decomposition is, $\pmb v = \overline{\langle \pmb{b_{1}}, \pmb v \rangle} \pmb{b_{1}} + \overline{\langle \pmb{b_{2}}, \pmb v \rangle} \pmb{b_{2}} + \dots + \overline{\langle \pmb{b_{n}}, \pmb v \rangle} \pmb{b_{n}} = \langle \pmb v, \pmb{b_{1}} \rangle \pmb{b_{1}} + \langle \pmb v, \pmb{b_{2}} \rangle \pmb{b_{2}} + \dots + \langle \pmb v, \pmb{b_{n}} \rangle \pmb{b_{n}}$, and therefore $v_{i} = \langle \pmb v, \pmb{b_{i}} \rangle, \ \overline{v_{i}} = \langle \pmb{b_{i}}, \pmb v \rangle$.

Using this, my first attempt at expanding becomes: $$ \begin{align} \langle \pmb u, \pmb v \rangle &= \bigg\langle u_{1} \pmb{b_{1}} + u_{2} \pmb{b_{2}} + \dots + u_{n} \pmb{b_{n}}, \ \pmb v \bigg\rangle \\ \\ &= u_{1} \langle \pmb{b_{1}},\pmb{v} \rangle + u_{2} \langle \pmb{b_{2}}, \pmb{v} \rangle + \dots + u_{n} \langle \pmb{b_{n}}, \pmb{v} \rangle \\ &= \sum_{i = 1}^{n} u_{i} \langle \pmb{b_{i}}, \pmb v \rangle \\ &= \sum_{i = 1}^{n} u_{i} \overline{v_{i}} \end{align} $$ So it's the same as my second method, and everything is resolved. However, my new questions are:

1. Is the new decomposition correct?
2. If it is, then how does the formula $\pmb v = \sum_{\pmb b \in B} \langle \pmb b,\pmb v \rangle \pmb b$, (found on this wiki page) become $\pmb v = \overline{\langle \pmb{b_{1}}, \pmb v \rangle} \pmb{b_{1}} + \overline{\langle \pmb{b_{2}}, \pmb v \rangle} \pmb{b_{2}} + \dots + \overline{\langle \pmb{b_{n}}, \pmb v \rangle} \pmb{b_{n}}$? Is this related to different inner product conventions between Maths/physics?

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EDIT 2: I attempted this from a different direction. Because $B = \{ \pmb {b_{1}}, \pmb {b_{2}}, \dots, \pmb {b_{n}} \}$ is a basis, it means that for every $\pmb v \in V$, there exist some unique scalars $v_{1}, v_{2},\dots,v_{n} \in F$ for which: $$\pmb v = v_{1} \pmb{b_{1}} + v_{2} \pmb{b_{2}} + \dots + v_{n} \pmb{b_{n}}$$Therefore, we have that: $$ \begin{align} \overline{\langle \pmb{b_{1}}, \pmb v \rangle} \pmb{b_{1}} + \overline{\langle \pmb{b_{2}}, \pmb v \rangle} \pmb{b_{2}} + \dots + \overline{\langle \pmb{b_{n}}, \pmb v \rangle} \pmb{b_{n}} &= \langle \pmb v, \pmb{b_{1}} \rangle \pmb{b_{1}} + \langle \pmb v, \pmb{b_{2}} \rangle \pmb{b_{2}} + \dots + \langle \pmb v, \pmb{b_{n}} \rangle \pmb{b_{n}} \\ &= \sum_{i=1}^n \langle \pmb v, \pmb{b_{i}} \rangle \pmb{b_{i}} \\ &= \sum_{i=1}^n \langle v_{1} \pmb{b_{1}} + v_{2} \pmb{b_{2}} + \dots + v_{n} \pmb{b_{n}}, \pmb{b_{i}} \rangle \pmb{b_{i}} \\ &= \sum_{i=1}^n \bigg( v_{1} \langle \pmb{b_{1}}, \pmb{b_{i}} \rangle \pmb{b_{i}} + v_{2} \langle \pmb{b_{2}}, \pmb{b_{i}} \rangle \pmb{b_{i}} + \dots + v_{n} \langle \pmb{b_{n}}, \pmb{b_{i}} \rangle \pmb{b_{i}} \bigg) \\ &= \sum_{i=1}^n v_{i} \langle \pmb{b_{i}}, \pmb{b_{i}} \rangle \pmb{b_{i}} = \sum_{i=1}^n v_{i} \pmb{b_{i}} \\ &= v_{1} \pmb{b_{1}} + v_{2} \pmb{b_{2}} + \dots + v_{n} \pmb{b_{n}} \\ &= \pmb v \\ \\ \text{Therefore we have: } & \pmb v = \sum_{i=1}^n \langle \pmb v, \pmb{b_{i}} \rangle \pmb{b_{i}} \end{align} $$ So, as is pointed out by @Arturo Magidin in the comments, this formula is correct for physics convention, but not for maths convention.

Answer 1

if $\pmb v = \langle \pmb{b_{1}}, \pmb v \rangle \pmb{b_{1}} + \langle \pmb{b_{2}}, \pmb v \rangle \pmb{b_{2}} + \dots + \langle \pmb{b_{n}}, \pmb v \rangle \pmb{b_{n}}$ then you would get \begin{align*} \langle \pmb{b_i}, \pmb v\rangle = \langle \pmb{b_i},\langle \pmb{b_{1}}, \pmb v \rangle \pmb{b_{1}}\rangle + \langle \pmb{b_i},\langle \pmb{b_{2}}, \pmb v \rangle \pmb{b_{2}}\rangle \pmb{b_{2}} + \dots + \langle \pmb{b_i},\langle \pmb{b_{n}}, \pmb v \rangle \pmb{b_{n}}\rangle = \langle \pmb{b_i}, \langle \pmb{b_i}, \pmb{v}\rangle \pmb{b_i} \rangle = \overline{\langle \pmb{b_i},\pmb{v}}\rangle \end{align*} which is wrong in general.

Can you guess the correct formula?