Can someone provide a proof for the following theorem and explain why $R$ is exactly the definition of restricting a linear mapping (operator) like $A$$V$ --> $V$, where $V$ is a finite-dimensional space, in an invariant subspace such as $S$; symbolically shown as $A|S$?
Theroem: Consider the mapping $A$ with the invariant subspace $S$. Then, for matrix $T$ constructed by the basis of $S$, there exists a square matrix $R$ such that $A T = T R$
I'm guessing that this is in a finite-dimensional space $V$, and $A$ is a linear operator from $V$ into $V$. Suppose $u_i, i=1\ldots n$ is a basis of the invariant subspace $S$. So for each $j$ you have $A u_j \in S$, therefore $A u_j = \sum_{i=1}^n c_{ij} u_i$ for some scalars $t_{ij}$. If $T$ is the matrix whose columns are $u_1, \ldots, u_n$, and $R$ the matrix with entries $c_{ij}$, that says $A T = T R$.