I am currently going through a book, "Tensor Algebra and Tensor Analysis for Engineers" by Itskov, and I am trying to understand what is meant by dual basis for a vector space. The author states that an arbitrary vector in a Euclidean space can be represented as:
$\textbf{x}=x^i\textbf{g}_i=x_i\textbf{g}^i$
In trying to understand this, I went to the Wikipedia page hoping for a concrete example, and it states that a dual basis for the Euclidean space of column vectors is the corresponding basis row vectors. But if the $\textbf{x}$ vector in the above equation is a column vector, and so the $\textbf{g}_i$ are basis column vectors, then how can it possibly be represented by a basis made of row vectors as in $\textbf{g}^i$? No combination of scalars times basis row vectors equals a column vector! If the Wikipedia article just has it wrong, then what is a good example to illustrate what the author saying with the above equation?
Link to Wiki:
The only reason why we represent elements in the dual as row vectors is so the product of elements in the dual with an element in the original space is given by matrix multiplication. If we don't care about this representation, there's no "row vector" and no "column vector." There's simply a vector.
In fact, if you look at the next page, it continues, for fixed $i,$ $$(x^j\mathbf{g}_j) \cdot \mathbf{g}^i = x^j\delta_j^i = x^i = \mathbf{x}\cdot \mathbf{g}^i \\ (x_j\mathbf{g}^j)\cdot \mathbf{g}_i = x_j\delta_i^j = x_i = \mathbf{x} \cdot \mathbf{g}_i$$ so they have the same coordinates. (Note: I've changed the order of the equalities to be more logical and to emphasize what the author is showing)