Eigenvalue of Matrix Powers

Theorem
Let $A$ be a square matrix.

Let $\lambda$ be an eigenvalue of $A$ and $\mathbf v$ be the corresponding eigenvector.

Then:
 * $A^n \mathbf v = \lambda^n \mathbf v$

holds for each positive integer $n$.

Proof
Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
 * $A^n \mathbf v = \lambda^n \mathbf v$

Basis for the Induction
$\map P 1$ is true, as this just says:
 * $A \mathbf v = \lambda \mathbf v$

which follows by definition of eigenvector.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $A^k \mathbf v = \lambda^k \mathbf v$

Then we need to show:
 * $A^{k + 1} \mathbf v = \lambda^{k + 1} \mathbf v$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \N_{> 0}: A^n \mathbf v = \lambda^n \mathbf v$