Properties of the pseudoinverse of a matrix Let $A $ be a $n \times m $ and matrix and let $B = (A^T A)^{-1} A^T $ be its pseudo inverse. 
If $A$ and $C$ satisfies 
$CAC = C$, $ACA = A$, $(CA)^T = CA$, and $(AC)^T = AC $, then can we conclude that $C$ is the pseudo inverse of $A$ ??
From $ACA = C$ we obtain $A = C (CA)^{-1} $ and so $ACA = C (CA)^{-1}$ taking transposes we obtain $(ACA)^T = (C (CA)^{-1} )^T \implies A^T AC = (CA)^{-T} C^T$. But I get stuck here. I dont see any way to express $C$ by itself. Any help would be greatly appreciated.
 A: For this to work, we need to assume, that $A^TA$ is invertible of course. That is the case, if the columns of $A$ are linearly independent.
If that is the case, then:
$(A^TA)C = A^T(AC) = (A^T(AC))^{TT} =((AC)^TA^{TT})^T =(ACA)^T = A^T$
So:
$C = (A^TA)^{-1}A^T$.
You can also prove, that:
$C = A^T(AA^T)^{-1}$
The other two conditions you gave are not needed to prove that. However, they make $C$ the unique matrix with these 4 properties. For if $C'$ is another matrix satisfying said properties, then:
$AC = (AC)^T = C^TA^T = C^T(AC'A)^T = C^TA^TC'^TA^T = (AC)^T(AC')^T = ACAC' = AC'$
and similarly $CA = C'A$. So:
$C = CAC = CAC' = C'AC' = C'$.
Does $C$ satisfy those 4 properties, if $C = (A^TA)^{-1}A^T$? Yes, it does:
We have: $CA = (A^TA)^{-1}A^T A = I$. Hence:
$CAC = C$
$ACA  = A$
$(CA)^T = CA$
and finally:
$(AC)^T = C^TA^T = ((A^TA)^{-1}A^T)^TA^T = (A^{TT}((A^TA)^{-1})^T)A^T = A(A^TA)^{-1}A^T = AC$
(Hetebrij already pointed this out in the comments. I'm not that fast ;) ).
A: By definition matrix $C$ is a (Moore–Penrose) pseudo-inverse of $A$ if 


*

*$ACA = A$

*$CAC = C$

*$\big(\,AC\,\big)^T = AC$

*$\big(\,CA\,\big)^T = CA$


Assume that $A$ has linearly independent columns, so that the matrix $A^TA$ is invertible.
Then 
$%the first item in definition of  Moore–Penrose inverse we have$
\begin{align}
(1) &\implies A^T = \big(\,ACA\,\big)^T = A^T \big(\,AC\,\big)^T\\
(2) &\implies A^T =  A^T \big(\,AC\,\big)^T = A^TAC \\
&\implies  \bbox[2.5pt, border:solid 1.5pt #F00000]{C = \big(\,A^TA\,\big)^{-1}A^T}\\
&&\blacksquare
\end{align}
Now, given $B:=\big(\,A^TA\,\big)^{-1}A^T$, let us check if it satisfies Moore–Penrose pseudo inverse criteria:
\begin{alignat}{2}
1.\quad& ABA = A\big(\,A^TA\,\big)^{-1}A^TA =A
\\
2.\quad& BAB = \big(\,A^TA\,\big)^{-1}A^T A\big(\,A^TA\,\big)^{-1}A^T = \big(\,A^TA\,\big)^{-1}A^T = B
\\
3.\quad& \big(\,AB\,\big)^T =  B^TA^T = A\,\Big(\big(\,A^TA\,\big)^{-1}\Big)^TA^T 
= A\Big(\big(\,A^TA\,\big)^{T}\Big)^{-1}A^T = A\big(\,A^TA\,\big)^{-1}A^T = AB
\\
4.\quad& \big(\,BA\,\big)^T = \Big(\big(\,A^TA\,\big)^{-1}A^T A\Big)^{T} = I^T = I = \big(\,A^TA\,\big)^{-1}A^T A = BA\\
&&\blacksquare
\end{alignat}
