TLDR; What is this thing: $\begin{bmatrix} \hat{x}&\hat{y}&\hat{z} \end{bmatrix}$?
I was thinking about basis vectors, and I wrote down a matrix product that I have not seen anywhere.
Suppose we have a position in $\mathbb{R}^3$ given by the vector $\vec{r}=x\hat{x}+y\hat{y}+z\hat{z}$.
This vector is commonly given as a column vector, $\vec{r}=\begin{bmatrix} x\\ y\\ z \end{bmatrix}$
What I noticed was that if we chose our nouns for the mathematical objects such that certain matrix multiplications "just work out", I can write another form:
$$ \vec{r}= \begin{bmatrix} \hat{x}&\hat{y}&\hat{z} \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} $$
What do I have to call the row vector so that this makes sense. I know it is not a "covector". Maybe it is a "fiber bundle" or some other "bundle"? I know in some notations, the "row vector of column vectors is a matrix", but I know there has to be a word for the "dual-covector-basisvector-matrix-thingy".
I did notice that with this form of writing vectors, the components of the form are "commutative" in the sense that "vector" is self-transpose or transpose invariant.
$$ \vec{r}= \begin{bmatrix} \hat{x}&\hat{y}&\hat{z} \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} x&y&z \end{bmatrix} \begin{bmatrix} \hat{x}\\ \hat{y}\\ \hat{z} \end{bmatrix} $$
The transpose of one form gives the other form:
$$ \begin{bmatrix} \hat{x}&\hat{y}&\hat{z} \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix}^T = \begin{bmatrix} x&y&z \end{bmatrix} \begin{bmatrix} \hat{x}\\ \hat{y}\\ \hat{z} \end{bmatrix} $$
Computing the matrix product between the two gives back the original component form of the vector.
$$ \vec{r} = \begin{bmatrix} \hat{x}&\hat{y}&\hat{z} \end{bmatrix} \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} x&y&z \end{bmatrix} \begin{bmatrix} \hat{x}\\ \hat{y}\\ \hat{z} \end{bmatrix} = x\hat{x}+y\hat{y}+z\hat{z} $$
With other work I have done, I do know that this is more general than just this specific application and can generalize out more.
It's just a matrix.
$$\begin{bmatrix} \boldsymbol{e}_{1} & \boldsymbol{e}_{2} & \boldsymbol{e}_{3} \end{bmatrix} =\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ So your product $$\begin{bmatrix} \boldsymbol{e}_{1} & \boldsymbol{e}_{2} & \boldsymbol{e}_{3} \end{bmatrix}\begin{bmatrix} x^{1}\\ x^{2}\\ x^{3} \end{bmatrix}=\begin{bmatrix} x^{1}\\ x^{2}\\ x^{3} \end{bmatrix}$$ Just reads as $$\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} x^{1}\\ x^{2}\\ x^{3} \end{bmatrix}=\begin{bmatrix} x^{1}\\ x^{2}\\ x^{3} \end{bmatrix}$$ Or more simply $$\mathbf{I}\boldsymbol x=\boldsymbol x$$ Your equation is just a statement of the identity transformation.
More technically speaking, this is cheating. The object $\begin{bmatrix} \boldsymbol{e}_{1} & \boldsymbol{e}_{2} & \boldsymbol{e}_{3} \end{bmatrix}$ is actually nonsense. The components of a tensor should be elements of a field. Vectors are not elements of a field.