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Problem

Find generalitazion for matrix A exponents, when $n\in\{1,2,3,\dots\},n \in \mathbb{N}$

$$A^n=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n , \text{ when } n \in \mathbb{N}$$

Proof generalization by induction.

Attempt to solve

By computing a set of $A$ exponent's $n\in \{\ 1,2,3,4 \}$. It is possible to form generalization that is applicable for set $n\in \{1,2,3,4 \}$

$$ A^1=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix},A^2=\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix},A^3=\begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix},A^4=\begin{bmatrix} 1 & 4 \\ 0 & 1 \end{bmatrix} \dots A^n =\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix} $$

Induction proof

Induction hypothesis

Assume expression is valid when $n=k$

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

Base case

When $n=1$ $$ A^1=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $$ which is valid by definition.

Induction step

When $n=k+1$

$$ A^{k+1} = \begin{bmatrix} 1 & k+1 \\ 0 & 1\end{bmatrix}$$

$$ A^{k+1}=A^kA^1 $$

By utilizing induction hypothesis we have

$$ \implies A^{k+1}=\begin{bmatrix} 1 & k \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} $$ By utilizing matrix multiplication we have $$ \implies A^{k+1} \begin{bmatrix} 1\cdot 1 + k\cdot 0 & 1 \cdot 1 + 1 \cdot k \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot 1 + 1 \cdot 1 \end{bmatrix} $$ $$ \implies A^{k+1}=\begin{bmatrix} 1 & k+1 \\ 0 & 1 \end{bmatrix} $$

$$ \tag*{$\square$} $$


EDIT

The point of posting this was to have comment on if my solution seems correct or not. If you can notice something that doesn't look right, let me know !

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marked as duplicate by Mark McClure, ArsenBerk, Chris Custer, user10354138, Lee David Chung Lin Nov 13 '18 at 2:00

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    $\begingroup$ It is correct I think. $\endgroup$ – ArsenBerk Nov 12 '18 at 19:35
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    $\begingroup$ It is correct, just that it is confusing the way you "define " $A$ in the first equation as you define $A$ to be what you later call $A^n$. $\endgroup$ – Surb Nov 12 '18 at 19:36
  • $\begingroup$ @Surb That would be typo. Should be fixed now $\endgroup$ – Tuki Nov 12 '18 at 19:41
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Your proof looks good to me. There is an alternative proof: Note that $A=I+N$, where $N=\pmatrix{0&1\\ 0&0}$ squares to zero. Therefore, by binomial expansion, $$ A^n=(I+N)^n=I+nN+\binom{n}{2}N^2+\cdots+\binom{n}{n}N^n=I+nN=\pmatrix{1&n\\ 0&1}. $$

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