# Proof matrix $A=\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}^n , \text{ when } n \in \mathbb{N}$ [duplicate]

## 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 !

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

• 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 $A^n$. – Surb Nov 12 '18 at 19:36
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, $$A^n=(I+N)^n=I+nN+\binom{n}{2}N^2+\cdots+\binom{n}{n}N^n=I+nN=\pmatrix{1&n\\ 0&1}.$$