Let $n \times n$-matrix A have precisely $k(1\leq k \leq n)$ stable eigenvalues(stable means the real part is strictly negative).
A full-rank $n \times k$-matrix $R_A^S$ is called a right stable matrix of $A$ if $$ AR_A^S=R_A^SS_-, $$ here $S_-$ is a $k \times k$ stable matrix(S only has stable eigenvalues.)
Clearly, the existence of $R_A^S$ can be deduced from Jordan canonical form. However, the right stable matrix for $A$ is not unique. For another right stable matrix $\widetilde{R}_A^S,$ such that $$ A\widetilde{R}_A^S=\widetilde{R}_A^S\widetilde{S}_-, $$
is it possible to deduce that $$ \widetilde{R}_A^S=R_A^S S_0, $$ where $S_0$ is an invertible $k\times k$-matrix?
Yes. By a change of basis, we may assume that $$ A=\pmatrix{B\\ &C} \quad\text{and}\quad R_A^S=\pmatrix{X\\ Y} $$ where $B$ is $k\times k$ stable, the eigenvalues of $C$ have non-negative real parts, and $X$ is $k\times k$. From $AR_A^S=R_A^SS$, we obtain $BX=XS$ and $CY=YS$. As all eigenvalues of $C$ have non-negative real parts and $S$ is stable, $I_k\otimes C-S^T\otimes I_{n-k}$ is invertible. Therefore the equation $CY=YS$ is satisfied by $Y=0$ only. Hence $R_A^S$ is in the form of $\pmatrix{X\\ 0}$ for some square matrix $X$, and $X$ must be invertible because $R_A^S$ has full column rank. For the same reason, $\widetilde{R}_A^S=\pmatrix{\widetilde{X}\\ 0}$ for some invertible matrix $\widetilde{X}$. Therefore $\widetilde{R}_A^S=R_A^SX^{-1}\widetilde{X}$.