about the Moore-Penrose inverse In this article  , I found this proof :

If $A$ is a partial isometry, we know that $A^*=A^{+}$and $A A^*$ is the orthogonal projection onto $\mathcal{R}\left(A A^*\right)=\mathcal{R}(A)$. Thus, $A A^* A=P_{\mathcal{R}(A)} A=A$. Now, $A^4=A A^2 A=A$ implies $A^2=A^*$.

which i don't understand why $A^4=A A^2 A=A$ implies $A^2=A^*$?
 A: I would expect there is an easier argument, but here is one.
Write $$P=A^*A,\qquad\qquad Q=AA^*.$$ We know that $$A^4=A,\qquad\qquad A=QAP.$$
Write $T=A^3$. We have
$$
PT=PA^3=A^*A^4=A^*A=P.
$$
Then $$P=PP^*=PTT^*P.$$
This we can write as
$$
P(1-TT^*)P=0.
$$
Since $TT^*\leq1$ (this follows from the fact that $\|A\|=\|A^*A\|^{1/2}=1$), we get that $(1-TT^*)^{1/2}P=0$, and so  $(1-TT^*)P=0$, which we may write as
$$
P=TT^*P
$$
Thus
$$
(1-Q)P=(1-Q)TT^*P=(1-Q)QTT^*P=0.
$$
It follows that $P=QP$, so $P\leq Q$.
Repeating all the above with the roles of $A$ and $A^*$ exchanged, we get that $Q\leq P$. So $Q=P$. From here we can take two paths.
The first path is to notice that now $A=PA=AP=QA=AQ$. Then$$
A^2=PA^2Q=A^*A^4A^*=A^*AA^*=A^*. 
$$
The second path is to notice that $P=Q$ is $
A^*A=AA^*$,
and hence $A$ is normal. From $A^4=A$ we know that $\sigma(A)\subset\{0,1,\omega,\omega^2\}$, where $\omega$ is a primitive root of unity. Therefore $A^3$ is a normal operator with $\sigma(A^3)\subset \{0,1\}$, so $A^3$ is selfadjoint (a projection, in fact). Now we have
$$
A(A^2)A=A,\qquad (A^2)A(A^2)=A^2,\qquad (A\,A^2)^*=A\,A^2,\qquad (A^2\,A)^*=A^2\,A.
$$
Then $A^2$ is a Moore-Penrose inverse for $A$ and by the uniqueness, $A^*=A^2$.
