Eigenvalue of Matrix Powers
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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:
- $\forall n \in \N_{> 0}: A^n \mathbf v = \lambda^n \mathbf v$
$\blacksquare$