# Given a matrix $A$ find $A^n$.

$A=$$\begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix} Find A^n. My input: A^2= \begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2\\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 4\\ 0 & 1 \end{bmatrix} A^3 = \begin{bmatrix} 1 & 6\\ 0 & 1 \end{bmatrix} ...... A^n = \begin{bmatrix} 1 & 2n\\ 0 & 1 \end{bmatrix} This was very basic approach. I want to know if there is any other way a smart trick or something to solve this problem ? • You could first propose A^n = \begin{bmatrix} 1 & 2n\\ 0 & 1 \end{bmatrix} from observation, then use induction I guess. Jul 29, 2018 at 12:37 • What you have done is a very good way to discover the answer. Now, you ought to prove the answer, and that can be done by Mathematical Induction. Jul 29, 2018 at 12:37 • But, @chítrungchâu, this matrix can't be diagonalized. Jul 29, 2018 at 12:38 • Thank you karn and gerry Jul 29, 2018 at 12:41 • Take into account that A=I+2N, where N is zero everywhere except for the top right corner entry. Since N^2=0, then A^n=(I+2N)^n=I^n+2nN+\text{ terms in which }N\text{ has higher degree}= I+2nN. – user578878 Jul 29, 2018 at 12:45 ## 3 Answers You can use this too$$A=\begin{pmatrix}1&2\\0&1\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}+\begin{pmatrix}0&2\\0&0\end{pmatrix}A=I_2+B$$And B is a nilpotent matrix \implies B^2=0$$A^n=(I_2+B)^n$$Use binomial theorem • Nice. Thank you Jul 29, 2018 at 12:47 • @Damn1o1 yw with binomilal identity you get the result directly since B^n=0 for n \ge 3 Jul 29, 2018 at 12:51 What you did was the smart approach. That is, you computed the first few terms of the sequence $$(A^n)_{n\in\mathbb N}$$ and you noticed a patern. The only thing that remains to be done is to prove that the pattern is real, but that's easy. Obviously,$$A^1=A=\begin{pmatrix}1&2\\0&1\end{pmatrix}$$and$$A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}\implies A^{n+1}=A.\begin{pmatrix}1&2n\\0&1\end{pmatrix}=\begin{pmatrix}1&2(n+1)\\0&1\end{pmatrix}.$$ The minimal polynomial of A is (x-1)^2 so for any entire function f we have$$f(A) = f(1)P + f'(1)Q$$for some P, Q polynomials in A. Plugging in f \equiv 1 and f(x) = x we get P = I and Q = A-I. Therefore for f(x) = x^n we have$$A^n = 1^n\cdot I + n1^{n-1}\cdot (A-I) = I + n(A-I) = \begin{bmatrix} 1 & 2n \\ 0 & 1\end{bmatrix}$\$

• That's a really nice solution @mechanodroid. The use of the minimal polynomial is really a nice one! One can obviously use induction or binomial as done in the previous posts, but this one is an absolute beauty. Really nice one! Jul 25, 2020 at 18:40