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It says in my notes that for eigenvector $v$ with eigenvalue $\lambda$ of matrix $A$, that $A^nv = \lambda^nv$, how does that work? I know that $Av = \lambda v$ but still...

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Keep multiplying the left by $A$.

If $A v = \lambda v$, then

$A^2 v = A \lambda v = \lambda Av = \lambda \cdot \lambda v = \lambda^2 v.$

And, in general, if $A^k v = \lambda^k v$, then

$A^{k+1} v = A \cdot A^k v = A \lambda^k v = \lambda^k Av = \lambda^k \cdot \lambda v = \lambda^{k+1} v.$

This proves the assertion by induction.

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Well, per definition you have that $A^{n}v = A(A(A...(A\cdot v)...))$. When you apply the equality $A\cdot v = \lambda v$ successively, you get it. For example, $A^{2}\cdot v = A(A\cdot v) = A(\lambda v) = \lambda (A\cdot v) = \lambda^{2}v$.

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Just use induction.

You got $Av=\lambda v$, so your equality holds if $n=1$.

Now assume it holds for $n$ and prove it for $n+1$. You have: $A^{n+1}v=A^n (Av)=A^n(\lambda v) = \lambda\ A^nv$ (because of the definition of $A^{n+1}$ and because of linearity) hence by inductive hypothesis $A^nv=\lambda^nv$ therefore: $A^{n+1}v=\lambda (\lambda^n v)=\lambda^{n+1}v$ as you wanted.