Let $M_1, M_2, M_3$ are $n\times n$ matrices with real entries, and modulus of the eigenvalues are strictly less than one. For $x,y$ are any $n\times 1$ vector, Could anyone tell me given a suitable norm can I say this norm-is less than something? $$\|M_1M_3M_2x- M_2M_1M_3y\|< ?$$
I know that the Matrix norm has some relation with its spectral-radius, from there we know that each of the matrix norms is $<1$. Thanks for any help.
The norm one can use Hilbert metric(norm) too if that works.
I'm not sure if this is the kind of result you are looking for, but if $M_2$ is invertible you can reason as follows.
Set $A:=M_1 M_3 M_2$ and $z:=M_2^{-1}y$. Then
\begin{align} M_1 M_3 M_2 x- M_2 M_1 M_3 y &= Ax - M_2 A z \\ &= (1-M_2) Ax + M_2 Ax - M_2 Az \\ &= (1-M_2) Ax + M_2 A(x-z) \end{align}
At this point you can use triangle inequality and your properties to infer
\begin{align} \Vert M_1 M_3 M_2 x- M_2 M_1 M_3 y \Vert \le & \,\Vert (1-M_2) \Vert \Vert x \Vert + \\ & + \Vert x-M_2^{-1}y \Vert \end{align}
NB I assumed the supremum norm, but any (submultiplicative) norm satisfying $\Vert AB x\Vert \le \Vert A\Vert \Vert B \Vert \Vert x\Vert$ would do