Prove a matrix Question by Induction

Question

I am struggling with this question. I get to to a certain point and then don't know what to do. So heres what I do.

Given:
A= $$ \begin{bmatrix} 2 & 0 \\ -1 & 1 \\ \end{bmatrix} $$

Proof: Consider $A^n$ = $$ \begin{bmatrix} 2^n & 0 \\ 1-2^n & 1 \\ \end{bmatrix} $$

(The Proposition)

(1) When n =1, $A^1$ = $$ \begin{bmatrix} 2^1 & 0 \\ 1-2^1 & 1 \\ \end{bmatrix} $$ = $$ \begin{bmatrix} 2 & 0 \\ -1 & 1 \\ \end{bmatrix} $$ Which is true. That is the proposition is true when n=1.

(2) Assume that the proposition is true when n=k

$A^k$ = $$ \begin{bmatrix} 2^k & 0 \\ 1-2^k & 1 \\ \end{bmatrix} $$

And i don't know what to do now. I started doing a K+1 proof but it went nowhere.

Answer 1

You just have to prove that the proposition is true for $n=k+1$, for that calculate $A^{k+1} = A^{k}\times A$ with using the fact that the proposition is true for $A^{k}$

Answer 2

This is the inductive step: Show that for all $k \geq 1$, we have $$A^{k+1} = A^kA = \left( \begin{matrix} 2^k & 0 \\ 1 - 2^k & 1 \end{matrix} \right)\left( \begin{matrix} 2 & 0 \\ -1 & 1 \end{matrix} \right) = \ \cdots $$

Fill in the blank by doing the matrix multiplication and show that the resulting matrix has the desired form for $A^{k+1}$.