# 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 $P \left({n}\right)$ be the proposition:

- $A^n \mathbf v = \lambda^n \mathbf v$

### Basis for the Induction

$P \left({1}\right)$ 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 $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ 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:

\(\displaystyle \lambda^{k+1} \mathbf v\) | \(=\) | \(\displaystyle \lambda \cdot \lambda^k \mathbf v\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \lambda A^k \mathbf v\) | Induction Hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle A^k \lambda \mathbf v\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle A^k A \mathbf v\) | Basis for the Induction | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle A^{k+1} {\mathbf v}\) |

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

Therefore:

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

$\blacksquare$