There are two ways to represent the same random variable:
$$\varepsilon_{1,t} \sim N(\vec{0},\Sigma\Sigma'), \text{ or } \Sigma\varepsilon_{2,t} \text{ where } \varepsilon_{2,t} \sim N(\vec{0},\mathbb{I}_n).$$
Consider the scalar term $\varepsilon_{1,t}'\delta'\delta\varepsilon_{1,t}$, and its equivalent $\varepsilon_{2,t}'\Sigma'\delta'\delta\Sigma\varepsilon_{1,t}$ where $\delta$ is a $(1 \times n)$ non-stochastic vector, and $\Sigma$ is $(n\times n)$ and also non-stochastic. Also $\Sigma\Sigma'$ is PSD. According to the properties of the quadratic form, $\mathbb{E}[\varepsilon_t'\Lambda\varepsilon_t]= \operatorname{tr}(\Lambda \Sigma^*) + \mu'\Lambda\mu$] where $\varepsilon_t \sim N(\mu,\Sigma^*)$ (doesn't need to be normal).
Assuming $\varepsilon_{1,t}=\Sigma\varepsilon_{2,t}$ does hold true, I get $\operatorname{tr}[\delta'\delta\Sigma\Sigma']=\operatorname{tr}[\Sigma'\delta'\delta\Sigma]$, since $\mu=\vec{0} \Rightarrow \mu'\Lambda\mu = 0$
Plugging arbitrary values into Matlab tells me this does not hold true.
Where does my logic fail?