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?