# Small eigenvalue question.

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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for example $A^2v = A(Av) = A(\lambda v) = \lambda\cdot Av = \lambda \cdot \lambda v = \lambda^2 v$. –  martini Mar 21 '12 at 13:33
Did you know that it is good form to accept answers that are correct and helpful to you? We can all see that you rarely do that and some people won't answer questions from people who do not. –  Graphth Mar 21 '12 at 13:37

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.

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