Properties of matrix equation CA=C I'm interested in the properties of the matric equation $CA=C$, where $C$ is an $m\times n$ matrix with real entries, and $A$ is a square $n\times n$ matrix with real entries, with $m\leq n$. What can I say about $A$, when $C$ is assigned? I know the following holds true:


*

*Let $A=C^*C$, where $C^*$ is any generalized inverse of $C$. Then
the system above always admits at least one solution.

*Whichever $C$, $A$ must be an orthogonal projection mapping the
rows of $C$ into themselves.


As for (1), can I go the other way round, i.e. $CA=C$ implies that $A$ be some linear function of (one of the) generalized inverses of $C$?
As for (2), does it imply that $A$ be symmetric (since real) and idempotent?
Thanks
 A: You have the equation $CA = C$, and you want to know what $A$ can be. If $C$ had a left inverse $D$, you could cancel it out and get $A = DC$.
However, your $C$ usually doesn't have a left inverse. The usual substitute when we would normally cancel something, but said operation doesn't make sense, is to bring everything over to once side and factor, so we get the equation
$$ C(A-I) = 0 $$
From this, we can immediately deduce a rather simple characterization of the possible values of $A-I$: each of its columns are right null vectors of $C$
A: See solving matrix equations $AX=B$ in
F.Gantmacher. Theory of matrices, AMS Chelsea publishing, 1959.
A: In fact you can only solve this question by this approach:
$CA=C$
$CA-C=0$
$C(A-I)=0$
Let $B=A-I$ ,
Then $CB=0$
Suppose $C=\left(\begin{array}{rrrr}c_{11}&c_{12}&\cdots&c_{1n}\\c_{21}&c_{22}&\cdots&c_{2n}\\\vdots&\vdots&\ddots&\vdots\\c_{m1}&c_{m2}&\cdots&c_{mn}\end{array}\right)$ and let $B=\left(\begin{array}{rrrr}b_{11}&b_{12}&\cdots&b_{1n}\\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\ddots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn}\end{array}\right)$ ,
Then $\left(\begin{array}{rrrr}c_{11}&c_{12}&\cdots&c_{1n}\\c_{21}&c_{22}&\cdots&c_{2n}\\\vdots&\vdots&\ddots&\vdots\\c_{m1}&c_{m2}&\cdots&c_{mn}\end{array}\right)\left(\begin{array}{rrrr}b_{11}&b_{12}&\cdots&b_{1n}\\b_{21}&b_{22}&\cdots&b_{2n}\\\vdots&\vdots&\ddots&\vdots\\b_{n1}&b_{n2}&\cdots&b_{nn}\end{array}\right)=\left(\begin{array}{rrrr}0&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{array}\right)$
$\left(\begin{array}{rrrr}\sum_{k=1}^nc_{1k}b_{k1}&\sum_{k=1}^nc_{1k}b_{k2}&\cdots&\sum_{k=1}^nc_{1k}b_{kn}\\\sum_{k=1}^nc_{2k}b_{k1}&\sum_{k=1}^nc_{2k}b_{k2}&\cdots&\sum_{k=1}^nc_{2k}b_{kn}\\\vdots&\vdots&\ddots&\vdots\\\sum_{k=1}^nc_{mk}b_{k1}&\sum_{k=1}^nc_{mk}b_{k2}&\cdots&\sum_{k=1}^nc_{mk}b_{kn}\end{array}\right)=\left(\begin{array}{rrrr}0&0&\cdots&0\\0&0&\cdots&0\\\vdots&\vdots&\ddots&\vdots\\0&0&\cdots&0\end{array}\right)$
$\therefore\begin{cases}\sum_{k=1}^nc_{1k}b_{k1}=0\\\sum_{k=1}^nc_{2k}b_{k1}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k1}=0\\\sum_{k=1}^nc_{1k}b_{k2}=0\\\sum_{k=1}^nc_{2k}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{1k}b_{kn}=0\\\sum_{k=1}^nc_{2k}b_{kn}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{kn}=0\end{cases}$
You should handle this system of linear equations.
