Let $v$ be an eigenvector of the matrix $A$. Show that $v$ is also an eigenvector of $A^3$. What is the corresponding eigenvalue?

Question

This is a question that I am trying to answer:

Let $\mathbf{v}$ be an eigenvector of the $n \times n$ matrix $A$. Show that $\mathbf{v}$ is also an eigenvector of $A^{3}$. What is the corresponding eigenvalue?

First of all I do not understand what is the notation $A^3$ is about? The only superscript notation of matrices that I know is for Transpose which is like $A^T$. What 3 is supposed to means, here?

Second this is how far I got with the proof: $$ A_{n,n} = \begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ a_{2,1} & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n,1} & a_{n,2} & \dots & a_{n,n} \end{pmatrix} $$ Now I need to calculate Eigenvalue of A: $$ \det(A - \lambda I) $$

But here I get stuck again! Because I do not know how I can show the determinant of $n \times n$ matrix. That would be something huge! Am I in the right track?

Answer 1

If $\mathbf{v}$ is an eigenvector of $A$, then there is an eigenvalue $\lambda$ such that $A\mathbf{v}=\lambda \mathbf{v}$, so $A^3\mathbf{v}=\lambda A^2\mathbf{v}=\lambda^2 A\mathbf{v} = \lambda^3 \mathbf{v}$.

Answer 2

In general if $v$ is an eigenvector of $A$, then $v$ is also an eigenvector of $A^n\ \forall\ n\ge 1$ the corresponding eigenvalue of $A^n$ is $\lambda^n$.

This is because $$Av=\lambda v\Rightarrow A^nv=A^{n-1}(Av)=A^{n-1}\lambda v=\cdots=\lambda^nv$$

Edited

Answer 3

Hint: $v$ being an eigenvector with corresponding eigenvalue $\lambda$ means $Av = \lambda v$. Now have a look at $A^3v$.