# prove the following property related to singular value decomposition

Suppose $A$ is a $n\times n$ matrix. Show that the following are equivalent:(i), $A^2=BA$ for some non-singular $B$. (ii) $rank(A)=rank(A^2)$. (iii), $$Range(A)\bigcap Ker(A)=\{0\}$$, (iv) there exits a non-singular $rank(A)\times rank(A)$ matrix $D$ and $n\times n$ non-singular matrix $P$ such that $$A=P\begin{pmatrix}D & 0\\ 0 & 0\end{pmatrix}P^{-1}$$

It's easy to prove $(i)\Longrightarrow(ii)\Longleftrightarrow(iii)$. But how is (iv) related to others? I can't figure out

By Fitting's lemma a square matrix over some field is similar to a matrix Dg$[D,N]$, where $D$ is non-singular and $N$ is nilpotent. Now it follows from (iii) that $N=0$. Furthermore, it is easy to show that (iii) follows directly from (iv), so then you must just prove (iii) implies (i) or (ii) implies (i).