The whole question looks like-
Let $A\in M_{n\times n} (\Bbb{C})$, show that the followings are equivalent-
(a) $A^2=BA$ for some non-singular matrix $B$
(b) $\text{rank}(A)=\text{rank}(A^2)$
(c) $\text{Im}(A)\cap \text{Ker}(A)=\{\theta\}$, where $\text{Im}(A)=\{AX|X\in\Bbb{C^n}\}$ and $\text{Ker}(A)=\{X\in\Bbb{C}^n|AX=\theta$}
(d) There exists non-singular matrices $P$ and $D$ of orders $n\times n$ and $\text{rank}(A)\times\text{rank}(A)$ respectively such that $$ A=P \begin{pmatrix} D & 0 \\ 0 & 0 \\ \end{pmatrix} P^{-1} $$
I want to prove the equivalance in the following manner- (a)$\implies$(b)$\implies$(c)$\implies$(d)$\implies$(a)
(a)$\implies$(b): We know $\text{rank}(BA)=\text{rank}(A)$ for some non-singular matrix $B$, hence $\text{rank}(A^2)=\text{rank}(A)$.
(b)$\implies$(c): Let, $AX\in\text{Im}(A)\cap\text{Ker}(A)\implies A(AX)=\theta\implies A^2 X=\theta\implies X\in \text{Ker}(A^2)$
$\dim(\Bbb{C^n})=\text{nullity}(A)+\text{rank}(A)$ and $\dim(\Bbb{C^n})=\text{nullity}(A^2)+\text{rank}(A^2)$
$\implies \text{nullity}(A)=\text{nullity}(A^2)$, since $\text{Ker}(A)\subseteq \text{Ker}(A^2)$, hence $\text{Ker}(A)=\text{Ker}(A^2)$.
Thus, we get $X\in \text{Ker}(A)\implies AX=\theta\implies \text{Im}(A)\cap\text{Ker}(A)=\theta$
Now, I cannot proceed further.
How to prove (c)$\implies$(d) and (d)$\implies$(a)? Can anybody suggest me a proper proof?
Thanks for assistance in advance.
For d implies a:
$$\begin{align} A^2 &= P\begin{pmatrix} D^2 & 0 \\ 0 & 0 \end{pmatrix} P^{-1} \\ &= P\begin{pmatrix} D & 0 \\ 0 & I \end{pmatrix} \begin{pmatrix} D & 0 \\ 0 & 0 \end{pmatrix} P^{-1} \\ &=P\begin{pmatrix} D & 0 \\ 0 & I \end{pmatrix} P^{-1} P \begin{pmatrix} D & 0 \\ 0 & 0 \end{pmatrix} P^{-1} \\ &=P\begin{pmatrix} D & 0 \\ 0 & I \end{pmatrix} P^{-1} A \end{align} $$ and $$ B = P\begin{pmatrix} D & 0 \\ 0 & I \end{pmatrix} P^{-1} $$ is non-singular
For c implies d, I'll leave out some details for you, but the idea is to let $k=\text{rank}(A)$ and $f_n,f_{n-1},\ldots,f_{n-k}$ be a basis for $\text{Ker}(A)$. Then we can extend this to a basis for all of $\mathbb{C}^n$ and let $P$ be the matrix that takes the standard basis to the new basis.