Let $$\mathbf{A} = \begin{bmatrix} 0 & i & 1 \\ 1 & 0 & i \\ i & 1 & 0 \end{bmatrix}$$ Find the limit $$\displaystyle{\lim_{n \to \infty} \det\mathbf{A}^n}$$ where $n \in \mathbb{N}$.
Does this limit exist? If we re-arrange the limit terms according to product of determinants rule [$\det \mathbf{A} \det \mathbf{B} = \det(\mathbf{AB})$], this yields
$$\displaystyle{\lim_{n \to \infty} \det\mathbf{A}^n} = {\displaystyle{\lim_{n \to \infty}} \displaystyle \prod_{n=1}^{\infty} \det\mathbf{A}}$$
Then, calculate the determinant of $\mathbf{A}$:
$$\det \mathbf{A} = 0 + 1 + i^3 - 0 - 0 - 0 = 1 +(-1)\cdot i = 1 -i$$
Now the limit is simplifed: $\displaystyle{\lim_{n \to \infty} (1-i)^n}$.
At this point, I'm stuck. Can we derive the subsequences with different limits which show that sequence limit does not exist here? WolframAlpha shows $1-\exp(2i [0 \space\text{to}\space\pi])$ as an answer, but I'm not able to decipher it.
Let $a_n:= \det A^n$. Then we have $|a_n|= \sqrt{2}^n$, hence $(a_n)$ is unbounded and therefore divergent.