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initial attempt failed, second attempt, let me calculate $\mathbb{E}[y^\top y]=\text{tr}\,\mathbb{E}[yy^\top]$ (using Isserlis theorem):

$$\mathbb{E}[y^\top y]=\sum_{ij}\mathbb{E}\bigl[A_{ij}\epsilon_j\epsilon_i\bigr]+\sum_{ijkl}\mathbb{E}\bigl[B_{ij} \epsilon_j \epsilon_l B_{kl} \epsilon_k\epsilon_i\bigr]$$ $$\qquad=\sum_{ij}A_{ij}\delta_{ij}+\sum_{ijkl}B_{ij}B_{kl}(\delta_{jl}\delta_{ki}+\delta_{ji}\delta_{lk}+\delta_{jk}\delta_{li})$$ $$\qquad=\text{tr}\,A+(\text{tr}\,B)^2+\text{tr}\,B^2+\text{tr}\,BB^\top.$$

let me calculate $\mathbb{E}[y^\top y]=\text{tr}\,\mathbb{E}[yy^\top]$ (using Isserlis theorem):

$$\mathbb{E}[y^\top y]=\sum_{ij}\mathbb{E}\bigl[A_{ij}\epsilon_j\epsilon_i\bigr]+\sum_{ijkl}\mathbb{E}\bigl[B_{ij} \epsilon_j \epsilon_l B_{kl} \epsilon_k\epsilon_i\bigr]$$ $$\qquad=\sum_{ij}A_{ij}\delta_{ij}+\sum_{ijkl}B_{ij}B_{kl}(\delta_{jl}\delta_{ki}+\delta_{ji}\delta_{lk}+\delta_{jk}\delta_{li})$$ $$\qquad=\text{tr}\,A+(\text{tr}\,B)^2+\text{tr}\,B^2+\text{tr}\,BB^\top.$$

initial attempt failed, second attempt, let me calculate $\mathbb{E}[yy^\top]$ (using Isserlis theorem):

$$\mathbb{E}[yy^\top]=\sum_{ij}\mathbb{E}\bigl[A_{ij}\epsilon_i\epsilon_j\bigr]+\sum_{ijkl}\mathbb{E}\bigl[B_{ij} \epsilon_j \epsilon_l B_{kl} \epsilon_k\epsilon_i\bigr]$$ $$\qquad=\sum_{ij}A_{ij}\delta_{ij}+\sum_{ijkl}B_{ij}B_{kl}(\delta_{jl}\delta_{ki}+\delta_{ji}\delta_{lk}+\delta_{jk}\delta_{li})$$ $$\qquad=\text{tr}\,A+(\text{tr}\,B)^2+\text{tr}\,B^2+\text{tr}\,BB^\top.$$

initial attempt failed, second attempt, let me calculate $\mathbb{E}[y^\top y]=\text{tr}\,\mathbb{E}[yy^\top]$ (using Isserlis theorem):

$$\mathbb{E}[y^\top y]=\sum_{ij}\mathbb{E}\bigl[A_{ij}\epsilon_j\epsilon_i\bigr]+\sum_{ijkl}\mathbb{E}\bigl[B_{ij} \epsilon_j \epsilon_l B_{kl} \epsilon_k\epsilon_i\bigr]$$ $$\qquad=\sum_{ij}A_{ij}\delta_{ij}+\sum_{ijkl}B_{ij}B_{kl}(\delta_{jl}\delta_{ki}+\delta_{ji}\delta_{lk}+\delta_{jk}\delta_{li})$$ $$\qquad=\text{tr}\,A+(\text{tr}\,B)^2+\text{tr}\,B^2+\text{tr}\,BB^\top.$$

For large $k$ the matrix $\epsilon\epsilon^\top\rightarrow \mathbb{E}[\epsilon\epsilon^\top]=I_k$, with fluctuations that vanish as $1/\sqrt k$, so $y\rightarrow (A+BB^\top)^{1/2}\epsilon$ and the elements of $y$ are Gaussian with zero mean and covariance matrix $C=A+BB^\top$.

initial attempt failed, second attempt, let me calculate $\mathbb{E}[yy^\top]$ (using Isserlis theorem):

$$\mathbb{E}[yy^\top]=\sum_{ij}\mathbb{E}\bigl[A_{ij}\epsilon_i\epsilon_j\bigr]+\sum_{ijkl}\mathbb{E}\bigl[B_{ij} \epsilon_j \epsilon_l B_{kl} \epsilon_k\epsilon_i\bigr]$$ $$\qquad=\sum_{ij}A_{ij}\delta_{ij}+\sum_{ijkl}B_{ij}B_{kl}(\delta_{jl}\delta_{ki}+\delta_{ji}\delta_{lk}+\delta_{jk}\delta_{li})$$ $$\qquad=\text{tr}\,A+(\text{tr}\,B)^2+\text{tr}\,B^2+\text{tr}\,BB^\top.$$