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

Question

I have a matrix:

$$ A=\begin{bmatrix}1 & a \\ 0 & 1\end{bmatrix} $$

which satisfies: $A\left( e_{1}\right)=e_{1},A\left( e_{2}\right)=ae_{1}+e_{2}, a\neq 0$

I need compute all powers $A^{n}, n\in \mathbb{Z}$ and find their matrices.

$$ A^n= \begin{bmatrix}1 & na \\ 0 & 1\end{bmatrix} $$

Is it true?

Is there a basis of $V$ (if $V$ is a 2-dimensional real vector space with basis $\left\{e_{1},e_{2}\right\}$) so that the matrix of $A$ is diagonal?

Answer 1

Write $$ A = I+ N = \pmatrix{1 & 0 \\ 0 & 1} + \pmatrix{ 0 & a \\ 0 & 0 }. $$ Then $N$ is nilpotent and $N^2=0$. Now $$ A^n = \sum_{k=0}^n {n\choose k} N^k = I+ {n\choose1} N = I + nN =\pmatrix{1&na\\0&1} , $$ where I used the convention $N^0=I$.

Answer 2

The formula is clearly true for $n=0$. Use induction for positive and negative $n$ to show that the formula is always true.

If the matrix was diagonalisable, then you would have $A = V \Lambda V^{-1}$ for some invertible $V$, where $\Lambda$ is a diagonal matrix of eigenvalues. However, in this case it is easy to compute the eigenvalues and derive a contradiction.

Addendum: My logic above could be clarified slightly. Since the eigenvalues of $A$ are all one, then we must have $\Lambda = I$ and hence $A=I$, a contradiction. Hence $A$ cannot be diagonalised.