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I have a triangular matrix $A$, and I need to find $A^{20}$.

$$ A = \begin{bmatrix} 3&1&0\\ 0&3&1\\ 0&0&3\\ \end{bmatrix} $$

I tried to use deagonalization method, but can't seem to find the correct eigen values and vectors.

Using $|A-\lambda I|=0$, I get $(3-\lambda)^3 = 0 \implies \lambda = 3$. So I'm guessing the eigen values all are 3?

For eigen vectors, using $|A-\lambda I|x=0$, I get $x_y = -x_z$. I'm not sure what to do next.

Lenny White
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    Binomial coefficients. You have $A = 3I + N$ The important thing is that $N$ and $3I$ commute, while $N$ is nilpotent: $N^3 = 0$ In general, when $BC = CB $ with square matrices, then $(B+C)^2 = B^2 + 2BC + C^2,$ then $(B+C)^3 = B^3 + 3 B^2 C + 3 B C^2 + C^3,$ and so on – Will Jagy May 02 '23 at 03:14
  • @WillJagy I see thank you! Would it mean however that I would need to expand and calculate a binomial to the power of 20? – Lenny White May 02 '23 at 03:22
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    You are missing the significance of $N^3 = 0$ – Will Jagy May 02 '23 at 03:26
  • Lenny White: "Would it mean however that I would need to expand and calculate a binomial to the power of 20?" Yes, that method ought to work well using the nilpotence of N. – Dan Asimov May 02 '23 at 05:32

1 Answers1

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This matrix is a Jordan block, it cannot be diagonalized.

For $k \ge 2$, we have $$A^k=\begin{bmatrix} 3^k & k\cdot 3^{k-1} & \binom{k}{2}3^{k-2} \\ 0 & 3^k & k \cdot 3^{k-1} \\ 0 & 0 & 3^k\end{bmatrix}$$

using the formula stated here.

Siong Thye Goh
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