I'm studying on Matrix Variate distributions by Gupta and Nagar. I can't figure out why the following identity holds: \begin{equation}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}=\text{tr}\{\Sigma^{-1}(X-M)\Psi^{-1}(X-M)'\}\tag{1}\end{equation} which is equation $(2.2.3)$, page 56. The notation has the following meaning:
- $X\in\mathbb{R}^{p\times n}$, $M\in\mathbb{R}^{p\times n}$;
- $\text{tr}\{\cdot\}$ is the trace function;
- $\Sigma\in\mathbb{R}^{p\times p}$, $\Psi\in\mathbb{R}^{n\times n}$ are symmetric and positively definite matrices;
- $\otimes$ is the Kronecker product;
- $\boldsymbol{x}\triangleq \text{vec}[X]$, $\boldsymbol{m}\triangleq \text{vec}[M]$, where $\text{vec}[X]\in\mathbb{R}^{pn}$ is the operator that stacks the columns $X_1$, $\dots$, $X_n$ of $X$ on top one each other, i.e. \begin{equation}X=\begin{bmatrix}X_1 & \dots & X_n\end{bmatrix}\Rightarrow \text{vec}[X]=\begin{bmatrix}X_1 \\ \vdots \\ X_n\end{bmatrix}\end{equation} same for $\text{vec}[M]=[M_1' \cdots M_n']'\in\mathbb{R}^{pn}$.
The authors get the identity, without specifying how, by using the following theorem (1.2.22, page 9): if $A\in\mathbb{R}^{p\times m}$, $B\in\mathbb{R}^{n\times q}$, $C\in\mathbb{R}^{q\times m}$, $D\in\mathbb{R}^{q\times n}$, $E\in\mathbb{R}^{m\times m}$, $X\in\mathbb{R}^{m\times n}$ then \begin{equation}\text{vec}[AXB]=(B'\otimes A)\text{vec}[X]\tag{2}\end{equation} \begin{equation}\text{tr}[CXB]=\text{vec}[C']'(I_q\otimes X)\text{vec}[B]\tag{3}\end{equation} \begin{equation}\text{tr}[DX'EXB]=\text{vec}[X]'(D'B'\otimes E)\text{vec}[X]=\text{vec}[X]'(BD\otimes E')\text{vec}[X]\tag{4}\end{equation} where $'$ denotes the transpose operator and $I_q$ is the $q\times q$ identity matrix.
Here is my first attempt to get the identity $(1)$: it's easy to see that $\text{vec}[\cdot]$ is linear, thus \begin{equation}\boldsymbol{x}-\boldsymbol{m}=\text{vec}[X-M]\end{equation} consequently from $(2)$, \begin{equation}(\Sigma^{-1}\otimes \Psi^{-1})\text{vec}[X-M]=\text{vec}[\Psi^{-1}(X-M)(\Sigma^{-1})']\end{equation} but $\Sigma$ is symmetric, thus $\Sigma^{-1}$ is symmetric, so \begin{equation}(\Sigma^{-1}\otimes \Psi^{-1})\text{vec}[X-M]=\text{vec}[\Psi^{-1}(X-M)\Sigma^{-1}]\end{equation} this means that the LHS of $(1)$ can be written as \begin{equation}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}=\text{tr}\{\text{vec}[\Psi^{-1}(X-M)\Sigma^{-1}]\text{vec}[X-M]'\}\end{equation} now for the general property of the trace $\text{tr}[AB]=\text{tr}[BA]$ holds \begin{equation}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}=\text{tr}\{\text{vec}[X-M]'\text{vec}[\Psi^{-1}(X-M)\Sigma^{-1}]\}\end{equation} and, for the the general property of the vectorization $\text{vec}[A]'\text{vec}[B]=\text{tr}[A'B]$, \begin{equation}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}=\text{tr}\{\text{tr}\{(X-M)'\Psi^{-1}(X-M)\Sigma^{-1}\}\}\end{equation} now, trivially, the trace of a scalar is the scalar itself, so \begin{equation}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}=\text{tr}\{(X-M)'\Psi^{-1}(X-M)\Sigma^{-1}\}\end{equation} and finally, once again for the property $\text{tr}[AB]=\text{tr}[BA]$, I conclude \begin{equation}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}=\text{tr}\{\Sigma^{-1}(X-M)'\Psi^{-1}(X-M)\}\end{equation} which is not $(1)$ because $(X-M)'$ is not in the right place.
Here is a second slimmer attempt. From $(4)$, by considering $D=I_n$ follows the identity \begin{equation}\text{tr}[X' E X B]=\text{vec}[X]'(B'\otimes E)\text{vec}[X]\end{equation} thus, by exploiting this formula, \begin{equation}\begin{aligned}\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})(\boldsymbol{x}-\boldsymbol{m})(\boldsymbol{x}-\boldsymbol{m})'\}&=\text{tr}\{(\Sigma^{-1}\otimes \Psi^{-1})\text{vec}[X-M]\text{vec}[X-M]'\}\\ &=\text{tr}\{\text{vec}[X-M]'(\Sigma^{-1}\otimes \Psi^{-1})\text{vec}[X-M]\}\\ &=\text{tr}\{\text{tr}\{(X-M)'\Psi^{-1}(X-M)(\Sigma^{-1})'\}\}\\ &=\text{tr}\{(X-M)'\Psi^{-1}(X-M)\Sigma^{-1}\}\\ &=\text{tr}\{\Sigma^{-1}(X-M)'\Psi^{-1}(X-M)\}\\ \end{aligned} \end{equation} which is the same result of the previous attempt.
It sound strange to me the hypothesis that $(1)$ contains a typo because in another book, Aspects of Multivariate Statistical Theory by Muirhead, it is expressed implicitly the identity $(1)$ (in equation $(1)$ of page 79). I don't think that Gupta and Nagar have merely copied an erroneous result from Muirhead.