I'm reading a paper and got stuck on one of the simplifications that was done without any elaboration. I've taken a course on Linear Algebra, but this is a little out of reach for me...

The simplification done is this (note that $T$ is not the transpose but a constant) where it's the second step I don't follow:

$$ L(\mathbf{Y}|\mathbf{A})\propto |\mathbf{A_0}|^T\exp\left[-0.5tr(\mathbf{ZA})^\prime(\mathbf{ZA})\right]\propto |\mathbf{A_0}|^T\exp\left[-0.5 vec(\mathbf{A})^\prime(\mathbf{I}\otimes \mathbf{Z}^\prime\mathbf{Z})vec(\mathbf{A})\right] $$ where $\mathbf{Z}=\begin{bmatrix} \mathbf{Y} & -\mathbf{X}\end{bmatrix}$ is $T\times (m+k)$ and $\mathbf{A}=\begin{bmatrix}\mathbf{A}_0 & \mathbf{A}_+\end{bmatrix}^\prime$ is $(m+k)\times m$. It looks quite trivial and it bothers me that I don't even know where to start.

I know that $tr(X^\prime Y)=vec(X)^\prime vec(Y)$, but that requires equal size of $X$ and $Y$, does it not? Otherwise $vec(X)^\prime vec(Y)$ wouldn't be possible. (I guess though that if the dimension of $X$ is $k\times p$ and the dimension of $Y$ is $m\times n$, if $kp=mn$ then it'd still work, but that's not the case here).

Can anyone point me in the right direction? It's from "Bayesian Methods for Dynamic Multivariate Models" by Christopher Sims and Tao Zha if anyone is interested in the source.


1 Answer 1


$\def\tr{\mathop{\rm tr}}\def\vec{\mathop{\rm vec}}$We have, as $ZA$ and $ZA$ have equal size \begin{align*} \tr\bigl((ZA)^tZA\bigr) &= \vec(ZA)^t\vec(ZA) \end{align*} Now note that $\vec(ZA) = (\mathrm{Id} \otimes Z)\vec(A)$, hence $\vec(ZA)^t = \vec(A)^t(\mathrm{Id}\otimes Z^t)$, givnig \begin{align*} \tr\bigl((ZA)^tZA\bigr) &= \vec(ZA)^t\vec(ZA)\\ &= \vec(A)^t(\mathrm{Id}\otimes Z^t)(\mathrm{Id} \otimes Z)\vec(A)\\ &= \vec(A)^t(\mathrm{Id} \otimes Z^tZ)\vec(A) \end{align*} as wanted.

  • $\begingroup$ Nice! Thank you. I read it as $[tr(ZA)^\prime](ZA)$ and not $tr[(ZA)^\prime(ZA)]$, so thanks for clearing that up too. $\endgroup$
    – hejseb
    Commented Jul 2, 2013 at 8:34
  • $\begingroup$ what is the vector $d$? $\endgroup$
    – zahraesb
    Commented May 12, 2022 at 17:30
  • $\begingroup$ Vector $d$? Id, as in Identity, denotes the identity operator. $\endgroup$
    – martini
    Commented May 12, 2022 at 18:04

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