Is this Matrix Normal Distribution correctly defined? In this paper, the following appears

It's stated that $V^{-1}$ is the row covariance matrix for the Matrix-Normal.
However, according to this link in wikipedia, and the references therein, and adapting their notation to the above, $V^{-1}$ should be the column covariance matrix instead.
Am I correct, or have I made a mistake? If so, I would like some help to understand where.
 A: Edit:
Regarding the Wiki definition (the second one), you can write the random matrix as
$$
X = U^{1/2}Z [V^{1/2}]^\intercal + M \tag{1},
$$
where $Z$ is a matrix of standard normals. When you take the $\text{vec}$ on both sides of that, you get
$$
\text{vec}(X) - \text{vec}(M) = \text{vec}(X-M) = (V^{1/2} \otimes U^{1/2}) \text{vec}(Z) .
$$
This is a column vector, and the density of this has the form of the standard vector-normal density. It has mean $\text{vec}(M)$, and covariance matrix
$$
(V^{1/2} \otimes U^{1/2}) (V^{1/2} \otimes U^{1/2}) ^\intercal = (V \otimes U).
$$
If you want to know the reason they call $U$ the row covariance matrix and $V$ the column covariance matrix, I think it helps if you figure all the columns are independent, and so you set $V$ equal to the identity. Then the covariance matrix is a block-diagonal $ I \otimes U$. You can convince yourself the block-diagonal form makes sense if you write $X$ in terms of its columns. Then taking the vec of it will stack all these columns on top of each other.
You can show the equivalence between these definitions using a few more properties of the kronecker product, the vec operator, and the trace operator:
\begin{align*}
\text{tr}[V^{-1}(X-M)^\intercal U^{-1} (X-M)] &= \text{tr}[(X-M)^\intercal U^{-1} (X-M)V^{-1}] \\
&= \text{vec}[X-M]^\intercal \text{vec}[U^{-1} (X-M)V^{-1}] \\
&=  \text{vec}[X-M]^\intercal (V^{-1} \otimes U^{-1}) \text{vec}[ X-M] \\
&=  \text{vec}[X-M]^\intercal (V \otimes U)^{-1} \text{vec}[ X-M].
\end{align*}
Looking at the second definition
\begin{align*}
\text{tr}[(\Gamma - M)^\intercal \Pi^{-1} (\Gamma - M)  V] &= \text{vec}(\Gamma - M)^\intercal \text{vec} [\Pi^{-1} (\Gamma - M)V]\\
&= \text{vec}(\Gamma - M)^\intercal (V \otimes \Pi^{-1})\text{vec}(\Gamma - M),
\end{align*}
so it's like they're thinking of
$$
X  = \Pi^{1/2} Z [V^{-1/2}]^\intercal + M
$$
or equivalently
$$
\text{vec}(X) - \text{vec}(M) = (V^{-1/2} \otimes \Pi^{1/2}) \text{vec}(M).
$$
So the confusion is understandable because yes, it appears the paper you're linking to has it backwards: $V^{-1}$ is the column-covariance matrix. Also, it adds to the confusion that they use the same letter as the other definition, but then invert it.
Old stuff I don't want to delete but might be pertinent:
I will also mention that there are two conventions used when defining normal random matrices. They both, however, make use of the definition of normal random vectors.
The other convention is to string out the rows of a matrix. That is not what's being done here, but I recall this was what was done in a multivariate statistics class I took a few years ago. This makes sense because data are often found in this format. Each row is independent of every other row. And the columns in each row are dependent in the same way. For example, see the bottom of page 52 in this book.
These ideas aren't totally separate. For example, if you want to use formulas with say Kronecker products that apply when you're using the traditional $\text{vec}$ operator, just take the transpose of your data matrix beforehand.
