Same Eigenvector to Transformation raised by nth power

Question

Why does $ T^nv=\lambda ^nv$ for an eigenvector $v\in V, \lambda\in \mathbb{F}$ and $T:V \to V$?
would appreciate an explanation how from $ Tv=\lambda v$ we get $ T^nv=\lambda ^nv$

Answer 1

Use induction. Clearly, for $n=1$, $Tv = \lambda v$. Now, assume for $n=k$ that $T^k v = \lambda^k v$. Then, $T^{k+1} v = T (T^k v) = T(\lambda^k v) = \lambda^k T v = \lambda^k (\lambda v) = \lambda^{k+1} v$. Thus, by induction, we see $T^n v = \lambda^n v$.

Answer 2

If you have $Tv = \lambda v$, then $T^2 v = T(Tv) = T(\lambda v) = \lambda (Tv) = \lambda(\lambda v) = \lambda^2 v$. You can use induction to get to $n$.