# 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:

 $\ds \lambda^{k + 1} \mathbf v$ $=$ $\ds \lambda \cdot \lambda^k \mathbf v$ $\ds$ $=$ $\ds \lambda A^k \mathbf v$ Induction Hypothesis $\ds$ $=$ $\ds A^k \lambda \mathbf v$ $\ds$ $=$ $\ds A^k A \mathbf v$ Basis for the Induction $\ds$ $=$ $\ds A^{k + 1} {\mathbf v}$

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

Therefore:

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

$\blacksquare$