Proof: Transposition of a matrix-normal random variable
Theorem: Let $X$ be an $n \times p$ random matrix following a matrix-normal distribution:
\[\label{eq:matn} X \sim \mathcal{MN}(M, U, V) \; .\]Then, the transpose of $X$ also has a matrix-normal distribution:
\[\label{eq:matn-trans} X^\mathrm{T} \sim \mathcal{MN}(M^\mathrm{T}, V, U) \; .\]Proof: For a random vector $X \in \mathbb{R}^n$ with probability density function $f_X(x)$, the probability density function of the invertible function $Y = g(X)$ is
\[\label{eq:pdf-invfct} f_Y(y) = \left\{ \begin{array}{rl} f_X(g^{-1}(y)) \, \left| J_{g^{-1}}(y) \right| \; , & \text{if} \; y \in \mathcal{Y} \\ 0 \; , & \text{if} \; y \notin \mathcal{Y} \end{array} \right.\]where $\lvert J_{g^{-1}}(y) \rvert$ is the determinant of the Jacobian matrix
\[\label{eq:jac} J_{g^{-1}}(y) = \left[ \begin{matrix} \frac{\mathrm{d}x_1}{\mathrm{d}y_1} & \ldots & \frac{\mathrm{d}x_1}{\mathrm{d}y_n} \\ \vdots & \ddots & \vdots \\ \frac{\mathrm{d}x_n}{\mathrm{d}y_1} & \ldots & \frac{\mathrm{d}x_n}{\mathrm{d}y_n} \end{matrix} \right]\]and $\mathcal{Y}$ is the set of possible outcomes of $Y$:
\[\label{eq:Y-range} \mathcal{Y} = \left\lbrace y = g(x): x \in \mathcal{X} \right\rbrace \; .\]In the present case, we have $Y = g(X) = X^\mathrm{T}$ and $X = g^{-1}(Y) = Y^\mathrm{T}$ and all $Y \in \mathcal{Y} = \mathbb{R}^{p \times n}$. For the vectorized matrices $X$ and $Y$, the Jacobian matrix is
\[\label{eq:XY-jac} J_{g^{-1}}(Y) = \left[ \begin{matrix} \frac{\mathrm{d}x_{11}}{\mathrm{d}y_{11}} & \frac{\mathrm{d}x_{11}}{\mathrm{d}y_{21}} & \ldots & \frac{\mathrm{d}x_{11}}{\mathrm{d}y_{pn}} \\ \frac{\mathrm{d}x_{21}}{\mathrm{d}y_{11}} & \frac{\mathrm{d}x_{21}}{\mathrm{d}y_{21}} & \ldots & \frac{\mathrm{d}x_{21}}{\mathrm{d}y_{pn}} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{\mathrm{d}x_{np}}{\mathrm{d}y_{11}} & \frac{\mathrm{d}x_{np}}{\mathrm{d}y_{21}} & \ldots & \frac{\mathrm{d}x_{np}}{\mathrm{d}y_{pn}} \end{matrix} \right] \in \mathbb{R}^{np} \; .\]Because by transposition, $y_{ji} = x_{ij}$, we have
\[\label{eq:dxij-dyji} \frac{\mathrm{d}x_{ij}}{\mathrm{d}y_{kl}} = \left\{ \begin{array}{rl} 1 \; , & \text{if} \; k = j \; \text{and} \; l = i \\ 0 \; , & \text{otherwise} \; . \end{array} \right.\]Thus, $J_{g^{-1}}(Y)$ is row-equivalent to $I_{np}$ and $\lvert J_{g^{-1}}(Y) \rvert = \lvert I_{np} \rvert = 1$. Therefore, we have:
\[\begin{split} f_Y(Y) &= f_X(g^{-1}(Y)) \, \left| J_{g^{-1}}(Y) \right| \\ &= \mathcal{MN}(Y^\mathrm{T}; M, U, V) \, \left| I_{np} \right| \\ &= \mathcal{MN}(Y^\mathrm{T}; M, U, V) \; . \end{split}\]The probability density function of the matrix-normal distribution is:
\[\label{eq:matn-pdf-X} f(X) = \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( V^{-1} (X-M)^\mathrm{T} \, U^{-1} (X-M) \right) \right] \; .\]Thus, substituting $X = Y^\mathrm{T}$, we get:
\[\label{eq:matn-pdf-Y-s1} f(Y) = \frac{1}{\sqrt{(2\pi)^{np} |U|^p |V|^n}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( V^{-1} (Y^\mathrm{T}-M)^\mathrm{T} \, U^{-1} (Y^\mathrm{T}-M) \right) \right] \; .\]Using $(A+B)^\mathrm{T} = (A^\mathrm{T} + B^\mathrm{T})$, we have:
\[\label{eq:matn-pdf-Y-s2} f(Y) = \frac{1}{\sqrt{(2\pi)^{np} |U|^p |V|^n}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( V^{-1} (Y-M^\mathrm{T}) \, U^{-1} (Y-M^\mathrm{T})^\mathrm{T} \right) \right] \; .\]Using $\mathrm{tr}(ABC) = \mathrm{tr}(CAB)$, we obtain
\[\label{eq:matn-pdf-Y-s3} f(Y) = \frac{1}{\sqrt{(2\pi)^{np} |U|^p |V|^n}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( U^{-1} (Y-M^\mathrm{T})^\mathrm{T} \, V^{-1} (Y-M^\mathrm{T}) \right) \right]\]which is the probability density function of a matrix-normal distribution with mean $M^T$, covariance across rows $V$ and covariance across columns $U$:
\[\label{eq:matn-pdf-Y-s4} f(Y) = \mathcal{MN}(Y; M^\mathrm{T}, V, U) \; .\]Metadata: ID: P144 | shortcut: matn-trans | author: JoramSoch | date: 2020-08-03, 22:21.