I would like to understand an article, which explains vectorial products in $n$ dimension.
As far as I know, the bases are vectors. If I multiply them with scalars, and add them together, then I can produce all the vectors in that vector-space.
But in this article, the bases are used as scalar values in a matrix. I don't understand how is it possible. Canonical base is maybe something else. I've found a Wikipedia article about it, but it says, that "it refers to the standard basis"
The relevant part of the article, which I can't understand:
There are given n-1 vectors $ a_i = (a_{i1}, \dots, a_{in}), i=1,\dots,n $ in $ \mathbb{R}^n $. We are looking for an $a_n \in \mathbb{R}^n $ which is perpendicular to all the others.
Let the coordinates of $a_n$ in the $e_1, \dots, e_n$ canonical base be $b_1, \dots, b_n$
In this case, it is clear that:
$$ a_n = \begin{vmatrix} e_1 & e_2 & \cdots & e_n \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n-1,1} & a_{n-1,2} & \cdots & a_{n-1,n} \end{vmatrix} = b_1 e_1 + \dots + b_n e_n $$
Note: I can not link the article, because it is not in english.
The determinant defining vector $a_n$ is simply an abuse of notation. It means that $a_n$ may be expressed as the cofactor expansion of the determinant of the "matrix"
$$ A = \left( \begin{matrix} e_1 & e_2 & \cdots & e_n \\ a_{11} & a_{12} & \cdots & a_{1n} \\ \cdots & \cdots & \cdots & \cdots \\ a_{n-1,1} & a_{n-1,2} & \cdots & a_{n-1,n} \end{matrix} \right) $$ along the first row, where the $e_i$ are vectors and the $a_{i,j}$ coefficients.
Thus, if we denote the cofactor matrix of $A$ by $\text{C}(A)$ and we write $a_n = b_1e_1 + \cdots + b_{n}e_{n}$ where $b_i$ are the real numbers defined as $$b_i := \text{C}(A)_{1,i} = \text{the (1, $i$)-th coefficient of the matrix $C(A)$}$$ then it follows from the Laplace expansion theorem that $$a_n\cdot a_i = 0 \quad \text{ if } \quad i \neq n $$ and $a_n$ is perpendicular to the $a_i$.