Is there any $n \times n$ matrix $A$ with complex entries such that $\operatorname{rank}A=2$, its eigenvalues are $\lambda,\lambda,0,0,\dots,0$ with $\lambda \neq 0$ and $X^2(X-\lambda)$ is its minimal polynomial?
My intuition is that there is not, since I can't find a specific example which satisfies these properties. However, I tried using the minimal and characteristic polynomials and also the Jordan form hoping that I could get something, but nothing came out useful in the end.
I am new to linear algebra and I am still learning, so an elementary solution (if there is one) for proving the non-existence would be much appreciated.
Later edit I added the condition that $X^2(X-\lambda)$ is $A's$ minimal polynomial.
You want a $2\times 2$ block with $X-\lambda$ as minimal polynomial and a $(n-2)\times(n-2)$ block with $X^2$ as minimal polynomial. The simplest matrix that satisfies these requirements is $$ \begin{pmatrix} \lambda & 0 &0 &\dotsb & 0 \\ 0 & \lambda &0&\dotsb & 0 \\ 0 & 0 &0&\dotsb & 1 \\ \vdots &\vdots &\vdots & \ddots & \vdots \\ 0&0&0&\dotsb&0 \end{pmatrix} $$ All non diagonal entries are zero, except $a_{3n}=1$.