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 $P \left({n}\right)$ 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:

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

$\blacksquare$