Show that the internal product can be fully described in terms of a given base through a given matrix.

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In my reference book, there is a demonstration about the positivity of an matrix A in an inner product, but there is a part that I didn't understand correctly. In the book it is as follows:

Let $V$ be a vector space of finite dimension over the body $\mathbb{F}$, which can be $\mathbb{R}$ or $\mathbb{C}$, provided with the inner product〈·, ·〉and $\beta = \{v_1 , ··· , v_n \}$ an ordered basis of $\mathbb{V}$ and $u,v \in V$, we can represent them as follows:

$$u=\sum_{i=0}^{n}x_jv_j\quad \text{and} \quad u=\sum_{i=0}^{n}y_iv_i$$

Thus, we have to:

$$〈u,v〉= 〈\sum_{i=0}^{n}x_jv_j\:,\:v〉=\sum_{i=0}^{n}x_j〈v_j\:,\:v〉=\underbrace{\sum_{i=0}^{n}x_j〈v_j\:,\:v〉\sum_{j=0}^{n}\bar y_i〈v_j\:,\:v_i〉}_{\huge{*}} =\sum_{i=0}^n \sum_{j=0}^{n} x_j\bar y_i〈v_j\:,\:v_i〉=Y^{*}AX$$

I really didn't understand what was done in ($\large{*}$) (the sum with respect to j). I know that $ \langle ku \, , \, \lambda v \rangle= k\bar \lambda \langle u \, , \,v \rangle$, and that way it's simpler to understand, but that product (it was put there to avoid problems with the summation) doesn't seem to be a multiplication by 1, because the base $\beta$ is not orthonormal.

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I think the confusion may lie in the fact that there are a few problems with the summation indices. Since $u, v \in V$ and $\beta = \left\{ v_1, \ldots, v_n \right\}$ is an ordered basis for $V$, then the summation index must start at $1$ and end at $n$. So, instead of what you have written, let $$u = \sum_{i=1}^{n} x_iv_i, \text{ and } v= \sum_{j=1}^{n} y_j v_j.$$ Make sure you do not confuse the above $v$ with any of the $v_1,\ldots,v_n \in \beta.$ Now for the computation. We have \begin{equation} \begin{split} \langle u, v \rangle &= \left\langle \sum_{i=1}^{n} x_iv_i, v \right\rangle \\ &= \sum_{i=1}^{n} x_i \langle v_i, v \rangle \\ &= \sum_{i=1}^{n} x_i \left< v_i, \sum_{j=1}^{n}y_jv_j \right> \\ &= \sum_{i=1}^{n} \sum_{j=1}^{n} x_i \langle v_i, y_jv_j \rangle \\ &= \sum_{i=1}^{n} \sum_{j=1}^{n} x_i \bar{y}_j \langle v_i, v_j \rangle . \end{split} \end{equation}

Does this clear things up?