Question about Column Space Matrix multiplication properties If say two square matrices A,B have the same column space, will it also hold that some multiplication of these matrices with another matrix will have the same column space?
i.e  if Col(A) = Col(B)
does Col(AC) = Col(BC)    for some other matrix C of the same dimensions.
It seems to make sense to me, but I couldn't work out how to prove it.
Thanks,
 A: If the two matrices have the same column space, it means that the corresponding linear maps have the same image. However they may not necessarily have the same kernel. So a counter example to the statement would be if $C$ maps all the vector to the kernel of $A$ but not for $B$.
E.g. take $$A=\begin{pmatrix} 0 & 1 \\ 0 &0 \end{pmatrix}$$
$$B=\begin{pmatrix} 1 & 0 \\ 0 &0 \end{pmatrix}$$
$$C=\begin{pmatrix} 1 & 0 \\ 0 &0 \end{pmatrix}$$
The map $A$ is just projection onto y axis and then rotate $90°$ clockwise, kernel of $A$ is any vector of the form $(x,0)$. $C$ is projection onto x axis, so it precisely maps every vector onto the kernel of $A$. As you can check $AC$ is the $0$ matrix, $BC=B$. They have different column space.
A: The counterexample @Bubububu has found is by taking a non-invertible $C$.
This wouldn't have occured with an invertible $C$.
Let us work with left multiplication by $C$ instead of right multiplication, because it is simpler (otherwise, transpose all). How can we prove that:
$$\tag{1}Col(A) = Col(B) \  \ \Longrightarrow  \ \ Col(CA) = Col(CB) $$
under the supplementary condition that $C$ is invertible ?
Some useful properties:

Lemma 1: Let $M_1 \cdots M_n$ be the columns of a certain matrix $M$;
$\sum_k a_k M_k = (M_1|\cdots|M_n)\begin{pmatrix}a_1\\\cdots\\a_n\end{pmatrix}=ML$ where $L$ is the column vector made of the coefficients of the linear combination.

as a corollary:

Lemma 2: $Col(M)$, being the set of all linear combinations of elements of $M$, is the (image) set of $M$, i.e., the set of all images $ML$ for any $M$.

as a corollary:

Lemma 3:  $Col(A)=Col(B) \Leftrightarrow \forall X, \exists L,L'  \ \text{such that} \   X=AL=BL'.$

With the help of this lemmas, (1) amounts to prove that
$$\forall Y, \exists L,L'  \ \text{such that} \  Y=CAL=CBL'$$
which is evident by multiplying by $C^{-1}$.
