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I'm trying to calculate $A^k$ for a 3x3 Jordan block matrix with 2 in the diagonal. I found this question in a previous exam for CS students, who were expected to solve it within 12 minutes at most.

At first, I attempted $A = VDV^{-1}$ to continue with $A^k = VD^kV^{-1}$. I found the triple Eigenvalue 2 but the corresponding Eigensystem only has dimension 1, so the matrix sadly is not diagonizable.

Given that, I'm not sure how I can solve this quickly. I would likely try to find the recursion formula for cells 1-1, 1-2 and 1-3 and prove each by induction. All other cells are symmetric (cells below the diagonal being 0 obviously). I recognize cell 1-1 is 2^k and hope the others are not too complicated, but this might be a gamble under time pressure.

Is there a different, better way?

mafu
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2 Answers2

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Write your Jordan block as $2 I + N$ where $N = \pmatrix{0 & 1 & 0\cr 0 & 0 & 1\cr 0 & 0 & 0\cr}$. Note that $N^2 = \pmatrix{0 & 0 & 1\cr 0 & 0 & 0\cr 0 & 0 & 0\cr}$ and $N^3 = 0$. So $$(2I+N)^k = 2^k I + k\; 2^{k-1} N + {k \choose 2} 2^{k-2} N^2 = \pmatrix{2^k & k\; 2^{k-1} & {k \choose 2} 2^{k-2}\cr 0 & 2^k & k\; 2^{k-1}\cr 0 & 0 & 2^k\cr}$$

Robert Israel
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I think I have gone over the 12 minutes, but you have to work out a few small cases, guess that $A^k=\left(\begin{array}{ccc}2^k & 2^{k-1}k & 2^{k-3}k(k-1)\\ 0 & 2^k & 2^{k-1}k \\ 0 & 0 & 2^k\end{array}\right)$ and then prove it by induction.

almagest
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