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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.

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  • $\begingroup$ Welcome to Mathematics Stack Exchange. Note that $(1-i)^8=16$ $\endgroup$ – J. W. Tanner Oct 7 '19 at 10:06
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    $\begingroup$ Consider the modulus. $\endgroup$ – xbh Oct 7 '19 at 10:07
  • $\begingroup$ @xbh If we take $n\mod 4$, our sequence will become -4, 16, -64, 256, ... Is this enough to make a conclusion, as our values fluctuate? $\endgroup$ – Floppy Drive Oct 7 '19 at 10:40
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Let $a_n:= \det A^n$. Then we have $|a_n|= \sqrt{2}^n$, hence $(a_n)$ is unbounded and therefore divergent.

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