Eigenvalue of Matrix Powers

Theorem
Let $A$ be a square matrix which is not nilpotent.

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

Then:

holds for each positive integer $n$.

Proof
We observe that a consequence of the definition of the eigenvalue of a square matrix $A$ is that the zero Matrix cannot have an eigenvalue.

Consider the zero matrix ${\mathbf 0_m}$ of order $m$.

Let $\mathbf v$ be a non-zero vector in $\R^m$.

Then $0_m \mathbf v = \mathbf 0$.

Now suppose that there exists $\lambda \in \R_{\ne 0}$ such that:
 * $0_m \mathbf v = \lambda \mathbf v$

Then:
 * $\lambda \mathbf v = \mathbf 0$

Hence $\mathbf v = \mathbf 0$, a contradiction.

The proof proceeds by induction.

Clearly, the statement holds for $n=1$.

Induction hypothesis:

Suppose that $A^n {\mathbf v} = \lambda^n {\mathbf v}$ holds for some positive integer $n$.

Then: