Matrix exponentiation -- Proof by induction

Question

I have been given that $M=\begin{bmatrix}1 & c\\-c & 1\end{bmatrix}.$

I have also been given that $ \det(M) = 0 ; $ thus, $ c= \pm i$.

How can I prove inductively that $M^n = 2^{n-1} M ? $

Answer 1

Let $P(n) : M^n = 2^{n-1}M$

$P(1): M^1 = 2^{0}M = 2^{1-1}M$ . So P(1) is true.

Let P(k) be true.

So, $P(k) : M^k = 2^{k-1}M$

$P(k+1):$

$M^{k+1} = M^k.M = 2^{k-1}M.M = 2^{k-1}M^2 = 2^{k-1}\begin{bmatrix}1 & c\\-c & 1\end{bmatrix}.\begin{bmatrix}1 & c\\-c & 1\end{bmatrix}$

$M^{k+1} = 2^{k-1} \begin{bmatrix}{1-c^2} & c+c\\-c-c & 1+1\end{bmatrix}$

As, $c = \pm i$,$c^2 = -1$

So, $M^{k+1} = 2^{k-1} \begin{bmatrix}{2} & 2c\\-2c & 2\end{bmatrix} = 2^{k-1}.2\begin{bmatrix}{1} & c\\-c & 1\end{bmatrix}= 2^{(k+1)-1}.M$

So, $P(k+1)$ is true.

Thus, $P(n)$ is true for all $n \in \Bbb Z^+$

Answer 2

This follows the suggestion by Saucy O'Path.

By the Cayley-Hamilton theorem, $(M-I)^2-I=0,$ so $\color{blue}{M^2=2M.}$

So we have proved a base case, $n=2$.

Now assume $M^{n-1}=2^{n-2}M $ for some $n$.

Multiply both sides by $M$ to get $M^n=2^{n-2}\color{blue}{M^2}=2^{n-2}\color{blue}{2M}=2^{n-1}M.$