If it's true for $\textrm{rank}(A)=\textrm{rank}(A^2)$

Question

I am trying to prove this and found this may be not true in general.Does anyone have any ideas about the counterexample for this statement?

score 3 · Accepted Answer · answered May 02 '16 at 02:08

3

$A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$

answered May 02 '16 at 02:08

Rene Schipperus

Thank you so much.Was preparing for the exam,lol – wenyuan wang May 02 '16 at 02:28
Good luck to you. – Rene Schipperus May 02 '16 at 02:31

1 Answers1