I'm a physics student studying linear algebra and the matrix representation of usual and dual vectors. I have several questions related to this topic, and I'm not sure if they make sense.
Firstly, the representation of usual vectors is done with column matrices for components and row matrices for bases. The inner product is
$||\vec{v}||^{2} = \begin{bmatrix} &v^{1}&\\[6pt] &v^{2}& \end{bmatrix}\begin{bmatrix} &v^{1}&\\[6pt] &v^{2}& \end{bmatrix}$
But I can't multiply two column matrices directly; I need to use the transpose of the first matrix, so that
$||\vec{v}||^{2} = \begin{bmatrix} &v^{1} &v^{2}& \end{bmatrix}\begin{bmatrix} &v^{1}&\\[6pt] &v^{2}& \end{bmatrix}= (v^{1})^{2}+(v^{2})^{2}$
For an orthonormal basis, I know that $V=V^{T}$. However, if my basis is not orthonormal, how can I correctly establish the relationship between:
$\begin{bmatrix} &v^{1}&\\[6pt] &v^{2}& \end{bmatrix} \longrightarrow \begin{bmatrix} &v^{1} &v^{2}& \end{bmatrix}$
since there is no transformation matrix that can turn a column vector into a row vector, I thought the following relation
$\begin{bmatrix} &v^{1}&\\ &v^{2}& \end{bmatrix}^{T} = \begin{bmatrix} &v^{1} &v^{2}& \end{bmatrix}\begin{bmatrix} &a &b&\\ &c & d& \end{bmatrix}$
where the $2\times2$ matrix would be something like a metric tensor. However, this seems arbitrary and doesn't convince me. Therefore, I would like to know a clear, step-by-step, and correct way to map a column vector to a row vector.
What you're asking is the following: given some inner product/bilinear form $\langle\cdot,\cdot\rangle$ on $\mathbb R^n$ and vectors $v, w \in \mathbb R^n$, how do we find a row matrix $v' \in \mathbb R^{1{\times}n}$ such that $$ v'w = \langle v, w\rangle? $$
Notice that we have a linear transformation $w \mapsto v'w$; we find the matrix of this transformation by evaluating on a basis, just like any other. So if $\mathbb e_i$ is the standard basis then $$ v' = \Bigl(\langle v, e_1\rangle,\, \langle v, e_2\rangle,\, \dotsc, \langle v, e_n\rangle\Bigr). $$ But we can do a bit better. Define the Gram matrix $G_{ij} = \langle e_i, e_j\rangle$ and notice that $$ v'w = \langle v, w\rangle = \langle v^ie_i, w^je_j\rangle = v^iG_{ij}w_j = v^TGw. $$ So we see immediately that $v' = v^TG$; we need only compute $G$ once and then we get the "modified transpose" $v'$ for all $v$.