1

Let $A$ be an $n$- square matrix such that $A^2= A$. Then how to prove that there exist a non-singular matrix $P\in M_n(\Bbb R)$ such that $P^{-1}AP$ is in normal form?

My approach: I have studied about this result: Suppose $A$ is an $m\times n$ matrix of rank $r$, then there exist two non-singular matrices $P$ (of size $m\times m$) and $Q$ (of size $n\times n$) such that $A=PNQ$ where $N=\pmatrix{I_r&0\\ 0&0}$ that is $I_r$ is the identity matrix of order $r$ and the $0$s here are null matrices.

I am not sure how to exploit this $A^2=A$ to solve this problem to proceed further. Thanks

Srijan
  • 12,862
  • 1
    See https://math.stackexchange.com/a/73875/111671 – obareey Feb 16 '20 at 10:00
  • The thing were I struggle without paper is to check, whether it is true that given regular square matrices $P,Q$ with $$QP = \begin{bmatrix} I_r & X\ Y & Z\end{bmatrix}$$ it already holds that $ X=Y=0$ and $Z = I_{n-r}$... – Jonas Linssen Feb 16 '20 at 10:03

0 Answers0