How to compute all powers $A^n$ and find is the matrix of $A$ diagonal?

Question

$V$ is a complex vector space with basis $\left\{e_{1},e_{2}\right\}$. Consider the linear operator on $V$ defined by $A\left( e_{1}\right)=ie_{1},A\left( e_{2}\right)=-ie_{1}, a\neq 0$. Compute all powers $A^{n}, n\in \mathbb{Z}$ and find their matrices. Is there a basis of $V$ so that the matrix of $A$ is diagonal?

Answer 1

As rank $A = 1$ you can check that $A^2 = \text{tr} A\times A = iA $ hence via induction: $$ A^n = i^{n-1} A $$

Finally, as we already have an eigenvector the last part is seay: in the basis $(e_1, e_1 + e_2$ the matrix of $A$ is \begin{pmatrix} i &0 \\ 0 & 0 \end{pmatrix}