Q:

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...

A:

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.

