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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 $A T = T R$

Saeed
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    You need to write this better. What is the context here? Hilbert space? Banach space? Finite-dimensional? Infinite-dimensional? You are mixing operators and matrices, and it is not clear why. – Martin Argerami Oct 20 '20 at 15:01
  • I modified my post. – Saeed Oct 20 '20 at 21:11
  • The context is finite-dimensional linear spaces, so mapping, operator, matrix are all equivalent. – Saeed Apr 09 '21 at 18:01

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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 $A T = T R$.

Robert Israel
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  • Thank you, Robert. I tried to edit your comment because t_ij is a typo and you actually meant to use c_ij as scalars. – Saeed Apr 09 '21 at 05:31