How does this transformation work?

Question

$$-\frac{N}{2}\ln|\Sigma|-\frac12 \sum_{n=1}^N (t_n-y_n)^T\Sigma^{-1}(t_n-y_n).$$

By rewriting the second term, we get

$$-\frac{N}{2}\ln|\Sigma|-\frac12 \operatorname{Tr}\left[\Sigma^{-1}\sum_{n=1}^N (t_n-y_n) (t_n-y_n)^T\right]$$

How does this transformation work?

Also, one more thing. What is $\Sigma$ if I was to set the derivative wrt to $\Sigma^{-1}$ for the above to 0.

Answer 1

Notice that $(t_n-y_n)^T\Sigma^{-1}(t_n-y_n)$ is a scalar, hence it is equal to its trace.

\begin{align} (t_n-y_n)^T\Sigma^{-1}(t_n-y_n) &= \operatorname{Tr}\left( (t_n-y_n)^T\Sigma^{-1}(t_n-y_n)\right) \\ &=\operatorname{Tr}\left( \Sigma^{-1}(t_n-y_n)(t_n-y_n)^T\right) \\ \end{align}

since trace of $AB$ is equal to trace of $BA$.

Also, notice that the trace of sum is equal to the sum of trace.