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I am looking at a paper which has the following manipulation:

$$ \exp \big(\sum_i(y_i - Ax_i)^T V^{-1}(y_i - Ax_i)\big) $$

Here $A$ is a matrix. $V$ is a symmetric, positive definite matrix (inverse of a covariance i.e. a precision matrix). The paper goes on to write this as:

$$ \exp\big(\operatorname{tr}(V^{-1}(Y-Ax)(Y-Ax)^T)\big) $$

However, it is not clear to me how this comes about! Can someone point me to what matrix identity has been used or guide to some proof of this?

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  • $\begingroup$ What are $x_i$ and $y_i$, are they vectors? What about $Y$ and $x$? $\endgroup$ – Pedro M. Feb 28 '15 at 2:44
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    $\begingroup$ It's a very common trick: if $x=a'Sb$ is a scalar, then $x=\text{Tr}(x)=\text{Tr}(a'Sb)=\text{Tr}(Sba')$. The first equality uses the fact that $x$ is a scalar and the last equality uses $\text{Tr}(AB)=\text{Tr}(BA)$ for all $A,B$ such that the matrix products make sense. $\endgroup$ – Kim Jong Un Feb 28 '15 at 2:58
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If $A\in\mathbb{R}^{n\times m}$ and $B\in\mathbb{R}^{m\times n}$ (and we do not assume $n=m$) then $\operatorname{trace}(AB)=\operatorname{trace}(BA)$.

First you have the scalar $$ \underbrace{(y_i - Ax_i)^T}\ \underbrace{V^{-1}(y_i - Ax_i)} $$ which is a product of matrices, but as a scalar it is its own trace. So it is equal to $$ \operatorname{trace}\left(\underbrace{V^{-1}(y_i - Ax_i)}\ \underbrace{(y_i - Ax_i)^T}\right). $$ Then the sum of the traces equals the trace of the sum, and the sum is $$ V^{-1}(Y-Ax)(Y-Ax)^T. $$

There is more on this in Wikipedia's article "Estimation of covariance matrices" (last time I looked).

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  • $\begingroup$ Also thanks for the wikipedia entry. That is a very useful page. $\endgroup$ – Luca Feb 28 '15 at 12:10
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If we stack the $x_i$s and $y_i$s column wise in a matrix, we get the matrix $X$ and $Y$ respectively.

And, $\sum_i(y_i - Ax_i)^TV^{-1}(y_i - Ax_i) = tr\left((Y - AX)^TV^{-1}(Y - AX)\right)$

Also, since the trace operator is invariant to cyclic permutations, $tr\left((Y - AX)^TV^{-1}(Y - AX)\right) = tr\left(V^{-1}(Y - AX)(Y - AX)^T\right)$

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