Notes

Eigenvalues of $A^k$

Suppose $\vec{v}$ I san eigenvector of A. $A\vec{v}=\lambda \vec{v}$

Multiply by A: $A^2\vec{v}=A\lambda \vec{v}=\lambda A\vec{v}=\lambda (\lambda \vec{v})={\lambda }^2\vec{v}$

So $\vec{v}$ is an eigenvector of $A^2$.

In general $A^k\vec{v}={\lambda }^k\vec{v}$

Suppose $A^{-1}$ then from$A\vec{v}=\lambda \vec{v}$ we get $\vec{v}=\lambda A^{-1}\vec{v}$

$A^{-1}\vec{v}=\frac{1}{\lambda }\vec{v}$

Diagonalization