Given two matrices $A\in\mathbb{R}^{3\times 3}$ and $B\in\mathbb{R}^{4\times 4}$, which don't have the same eigenvalues. Suppose there exists a matrix $C\in\mathbb{R}^{4\times 3}$ satisfying $CA=BC$. If the matrix $A$ is given, and its eigenvalues are known, how to find the matrix $C$ without knowing the detailed entries of $B$?
I was wondering what does the condition "don't have the same eigenvalues" mean. We won't use the definition of inverse matrix of a non-square matrix.
Let $C$ be the solution to the matrix equation $XA=BX$.
By mathematical induction, $XA^{n}=B^{n}X,n\geq1$. Set $f(x)$ the characteristic polynomial of matrix $A$, by Hamilton-Cayley theorem, $f(A)=O$. Since $f(A)$ is the combination of $I, A, A^2,\ldots,A^{n}$ and it is finite, so $Xf(A)=f(B)X$.
$A$ and $B$ don't have the same eigenvalue, set the eigenvalues of $A$ to $\lambda_{1},\ldots,\lambda_{n}$, then the roots of polynomial $f(x)$ are not as same as the eigenvalues of $B$, which derives the result $f(B)\neq O$. It's non-singular, because if it's singular, then $\vert f(B)\vert=\vert(B-\lambda_{1}I)^{m_{1}}\cdots(B-\lambda_{n}I)^{m_{n}}\vert=0$, meaning $A$ and $B$ have at least one same eigenvalue which leads to a contradiction. So $f(B)$ has an inverse, $f(B)X=O\Longrightarrow X=O$. The matrix equation has only one solution $C=O$.