How to show that $T^n = A^n = A$ for an idempotent matrix $A$

Question

suppose $A$ is a $2\times 2$ matrix given as

$$\pmatrix {3 & −6\cr 1 & -2}$$

Find the linear map $T : {\bf R}^2 \to{\bf R}^2$ associated to $A$ and find $T^{5001}$ based on its associated matrix.

I know that $A$ is an idempotent matrix but how do I solve this question with a general proof?

Answer 1

I suppose what you mean by "general proof" is to use induction to solve this.

So maybe first write down the statement you want to show: In this case it is $P_n:A^n=A$ for $n\in\mathbb{N}$.

So first you show that the base case is true, that is when $n = 1$. This is trivial.

Then you show that assuming $P_k$ is true for some $k\in \mathbb{N}$, i.e. $A^k=A$, then $P_{k+1}$ is true $(A^{k+1}=A)$.

Therefore $A^n=A$ $\forall n\in \mathbb{N}$.