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$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 ?

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  • $\begingroup$ You could first propose $A^n = \begin{bmatrix} 1 & 2n\\ 0 & 1 \end{bmatrix}$ from observation, then use induction I guess. $\endgroup$
    – Karn Watcharasupat
    Jul 29, 2018 at 12:37
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    $\begingroup$ 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. $\endgroup$
    – Gerry Myerson
    Jul 29, 2018 at 12:37
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    $\begingroup$ But, @chítrungchâu, this matrix can't be diagonalized. $\endgroup$
    – Gerry Myerson
    Jul 29, 2018 at 12:38
  • $\begingroup$ Thank you karn and gerry $\endgroup$
    – Daman
    Jul 29, 2018 at 12:41
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    $\begingroup$ Take into account that $A=I+2N$, where $N$ is zero everywhere except for the top right corner entry. Since $N^2=0$, then $A^n=(I+2N)^n=I^n+2nN+\text{ terms in which }N\text{ has higher degree}= I+2nN$. $\endgroup$
    – user578878
    Jul 29, 2018 at 12:45

3 Answers 3

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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

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  • $\begingroup$ Nice. Thank you $\endgroup$
    – Daman
    Jul 29, 2018 at 12:47
  • $\begingroup$ @Damn1o1 yw with binomilal identity you get the result directly since $B^n=0$ for $n \ge 3$ $\endgroup$
    – user577215664
    Jul 29, 2018 at 12:51
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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$$A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}\implies A^{n+1}=A.\begin{pmatrix}1&2n\\0&1\end{pmatrix}=\begin{pmatrix}1&2(n+1)\\0&1\end{pmatrix}.$$

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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

$$A^n = 1^n\cdot I + n1^{n-1}\cdot (A-I) = I + n(A-I) = \begin{bmatrix} 1 & 2n \\ 0 & 1\end{bmatrix}$$

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    $\begingroup$ 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! $\endgroup$
    – Ralph Clausen
    Jul 25, 2020 at 18:40

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