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
See solving matrix equations $AX=B$ in

F.Gantmacher. Theory of matrices, AMS Chelsea publishing, 1959.
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 $C=\left(\begin{array}{rrrr}c_{11}&amp;c_{12}&amp;\cdots&amp;c_{1n}\\c_{21}&amp;c_{22}&amp;\cdots&amp;c_{2n}\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\c_{m1}&amp;c_{m2}&amp;\cdots&amp;c_{mn}\end{array}\right)$ and let $B=\left(\begin{array}{rrrr}b_{11}&amp;b_{12}&amp;\cdots&amp;b_{1n}\\b_{21}&amp;b_{22}&amp;\cdots&amp;b_{2n}\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\b_{n1}&amp;b_{n2}&amp;\cdots&amp;b_{nn}\end{array}\right)$ ,

Then $\left(\begin{array}{rrrr}c_{11}&amp;c_{12}&amp;\cdots&amp;c_{1n}\\c_{21}&amp;c_{22}&amp;\cdots&amp;c_{2n}\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\c_{m1}&amp;c_{m2}&amp;\cdots&amp;c_{mn}\end{array}\right)\left(\begin{array}{rrrr}b_{11}&amp;b_{12}&amp;\cdots&amp;b_{1n}\\b_{21}&amp;b_{22}&amp;\cdots&amp;b_{2n}\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\b_{n1}&amp;b_{n2}&amp;\cdots&amp;b_{nn}\end{array}\right)=\left(\begin{array}{rrrr}0&amp;0&amp;\cdots&amp;0\\0&amp;0&amp;\cdots&amp;0\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\0&amp;0&amp;\cdots&amp;0\end{array}\right)$

$\left(\begin{array}{rrrr}\sum_{k=1}^nc_{1k}b_{k1}&amp;\sum_{k=1}^nc_{1k}b_{k2}&amp;\cdots&amp;\sum_{k=1}^nc_{1k}b_{kn}\\\sum_{k=1}^nc_{2k}b_{k1}&amp;\sum_{k=1}^nc_{2k}b_{k2}&amp;\cdots&amp;\sum_{k=1}^nc_{2k}b_{kn}\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\\sum_{k=1}^nc_{mk}b_{k1}&amp;\sum_{k=1}^nc_{mk}b_{k2}&amp;\cdots&amp;\sum_{k=1}^nc_{mk}b_{kn}\end{array}\right)=\left(\begin{array}{rrrr}0&amp;0&amp;\cdots&amp;0\\0&amp;0&amp;\cdots&amp;0\\\vdots&amp;\vdots&amp;\ddots&amp;\vdots\\0&amp;0&amp;\cdots&amp;0\end{array}\right)$

$\therefore\begin{cases}\sum_{k=1}^nc_{1k}b_{k1}=0\\\sum_{k=1}^nc_{2k}b_{k1}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k1}=0\\\sum_{k=1}^nc_{1k}b_{k2}=0\\\sum_{k=1}^nc_{2k}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{k2}=0\\\vdots\\\sum_{k=1}^nc_{1k}b_{kn}=0\\\sum_{k=1}^nc_{2k}b_{kn}=0\\\vdots\\\sum_{k=1}^nc_{mk}b_{kn}=0\end{cases}$

You should handle this system of linear equations.