If $X=\left(\begin{matrix}1 & -\frac{1}{5}-\frac{2}{5}i \\ 2+i & -i \end{matrix}\right)$, find the modulus of the sum of the elements of $X^{2023}$

Question

If $X=\left(\begin{matrix}1 & -\frac{1}{5}-\frac{2}{5}i \\ 2+i & -i \end{matrix}\right)$, determine the absolute value of the sum of the elements of $X^{2023}$

I was recently given this problem in an exam, and it completely stumped me. I am really unsure how to approach it. What I tried to no avail is to raise $X$ to a few powers and see if I can spot a general formula, but there didn't seem to be any that I could notice. Next I noticed $\det(X)=1$, so I attempted to apply the Cayley-Hamilton relation: $$X^2 - (1-i)X + I_2 = 0_2$$ Then I tried to successively multiply the whole equation by $X$ and replace any $X$s raised to the second power with $X^2 = (1-i)X-I_2$, but I couldn't find any recurrent relationship here either.

Any ideas how this would be solved any other way? Or perhaps there were helpful relations that I wasn't able to find in the two methods I tried?

Answer 1

You're on the right track, but your computation of the determinant is incorrect.

Hint We have $$\det X = (1)(-i) - \left(-\frac{1}{5} - \frac{2}{5} i\right)(2 + i) = (-i) - (-i) = 0 .$$

As you write, $\operatorname{tr} X = 1 - i$, so Cayley-Hamilton implies that $X^2 - (1 - i) X + 0 = 0$, i.e., that $$X^2 = (1 - i) X .$$ It's straightforward to find a recurrence formula for $X^k$ and hence for $\left\vert\sum_{j, k} (X^m)_{jk}\right\vert$.