Given matrix A with eigenvalue $\lambda$ and corresponding eigenvector x, prove $A^k$ has eigenvalue $\lambda^k$ Given matrix A with eigenvalue $\lambda$ and corresponding eigenvector x, prove $A^k$ has eigenvalue $\lambda^k$ for the same eigenvector x for any positive integer k.
Can I just use the eigenvalue definition Ax = $\lambda$x, divide x on both sides, then raising the power?

1.)  $Ax = \lambda x$, given.
2.)  Inductive hypothesis:   $A^kx = \lambda^k x$ for some positive integer $k$.
3.)  Operate on the equation of (2) with $A$:  $A^{k + 1}x = A(A^k x) = A(\lambda^k)x = \lambda^k (Ax) = \lambda^{k + 1} x$.
4.)  Conclude that $A^k x = \lambda^k x$ for all positive integers $k$.
QED.
Note Added in Edit, Tuesdqy 2 August 2022 8:05 PM PST:  In these remarks we shall extend and clarify the above , which we observe binds for $\lambda \in \Bbb F$ and $A \in M_{n \times n}(\Bbb F)$, $\Bbb F$ an arbitrary field.  We thus have, for $\alpha, \beta \in \Bbb F$, 1.) $\alpha A^kx = \alpha \lambda^kx$, whence 2.) $(\alpha A^k + \beta A^l)x = \alpha A^kx + \beta A^lx = \alpha \lambda^kx + \beta \lambda^lx = (\alpha \lambda^k + \beta \lambda^l)x$, indicating that 3.) $p(A)x = p(\lambda)x$ for any $p(y) \in F[y]$; finally, 4.) if the field $\Bbb F$ is possessed of a norm $\vert \cdot \vert: \Bbb F \to \Bbb R$ under which it is closed in the sense that Cauchy sequences converge to values (points) in $\Bbb F$ (as is the case with $\Bbb R$ or $\Bbb C$), then there is a legitimate context for considering convergent sequences and series of powers of $A$;  End of Note.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
