6
$\begingroup$

$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 ?

$\endgroup$
7
  • $\begingroup$ You could first propose $A^n = \begin{bmatrix} 1 & 2n\\ 0 & 1 \end{bmatrix}$ from observation, then use induction I guess. $\endgroup$ Jul 29, 2018 at 12:37
  • 1
    $\begingroup$ 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. $\endgroup$ Jul 29, 2018 at 12:37
  • 1
    $\begingroup$ But, @chítrungchâu, this matrix can't be diagonalized. $\endgroup$ Jul 29, 2018 at 12:38
  • $\begingroup$ Thank you karn and gerry $\endgroup$
    – Daman
    Jul 29, 2018 at 12:41
  • 1
    $\begingroup$ 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

15
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ 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 $n \ge 3$ $\endgroup$ Jul 29, 2018 at 12:51
7
$\begingroup$

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$$A^n=\begin{pmatrix}1&2n\\0&1\end{pmatrix}\implies A^{n+1}=A.\begin{pmatrix}1&2n\\0&1\end{pmatrix}=\begin{pmatrix}1&2(n+1)\\0&1\end{pmatrix}.$$

$\endgroup$
4
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ 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! $\endgroup$ Jul 25, 2020 at 18:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.