# covariant and contravariant form of a matrix

I'm following a paper to solve this equation:

$y_{j}=y_{o}$ + A$\eta^{T}$ (Eq. 2)

My question is about the term $\eta^{T}$. In the paper says:

"With symbol $\eta$, we denoted a 1 × 6 contravariant matrix with each element being a random number from the interval (0, 1) and $\eta^{T}$ , figuring in Eq.(2), is its covariant form."

I'm not sure If I understood correctly what is $\eta^{T}$ .

Is it basically a matrix 6 x 1, with the elements being a random number from the interval (0, 1) ? Something like that:

$\eta^{T}$= $\left[ {\begin{array}{cc} 0.548 \\ 0.365 \\ 0.889 \\ 0.725 \\ 0.427 \\ 0.169 \\ \end{array} } \right]$

Just in case, this is the paper: http://articles.adsabs.harvard.edu/pdf/2017CoSka..47....7N

(In Pag. 6)

Note:

A is a 6x6 covariance matrix

$y_{j}$ is a 6x1 matrix

$y_{o}$ is a 6x1 matrix

Thanks in advance for any help

$g_{ij} \eta^j=\eta_i \,,$
where $\eta^j \rightarrow \eta$ and $\eta_i \rightarrow \eta^T$ In the paper this is not clear, it seems that the covariant tensor is just the matrix transpose of the contravariant one. It looks like just jargon. There is some metric in that 6 dimensional space?.