Eigenvectors and Eigenvalues - cubic equation?

Question

So, I have the fact that : Av = λv

Where λ is the eigenvalue. I tried substituting in v=A^-1*v* λ to the equation but this didn't get me anywhere. Any help???

Answer 1

Let $\vec{v}$ be an eigenvector of $A$ with eigenvalue $\lambda$, meaning that we have $A \vec{v} = \lambda \vec{v}$.

Claim: $\vec{v}$ is an eigenvector of $A^3 - 7 A^2 + I$ with eigenvalue $\lambda^3 - 7 \lambda^2 + 1$.

Proof:

• First we notice that $A^2 \vec{v} = A (A \vec{v}) = A (\lambda \vec{v}) = \lambda (A \vec{v}) = \lambda ( \lambda \vec{v}) = \lambda^2 \vec{v}$. Thus, $\vec{v}$ is an eigenvector of $A^2$ with eigenvalue $\lambda^2$. If you do the same computation for $A^3$ instead of $A^2$, then you will find out that $\vec{v}$ is an eigenvector of $A^3$ with eigenvalue $\lambda^3$, i.e. $A^3 \vec{v} = \lambda^3 \vec{v}$.
  (Note: In general, $\vec{v}$ is an eigenvector of $A^k$ with eigenvalue $\lambda^k$, i.e. $A^k \vec{v} = \lambda^k \vec{v}$ where $k$ is a non-negative integer.)
• With that knowledge we obtain $$(A^3 - 7 A^2 + I)\vec{v} = A^3 \vec{v} - 7 A^2 \vec{v} + I \vec{v} = \lambda^3 \vec{v} - 7 \lambda^2 \vec{v} + \vec{v} = (\lambda^3-7\lambda^2+1)\vec{v},$$ so it turns out that $\vec{v}$ is an eigenvector of $A^3 - 7 A^2 + I$ with eigenvalue $\lambda^3-7\lambda^2+1$ which was our claim.