The matrix analogue of the Theorem about diagonalizable operators - Linear Algebra by Hoffman and Kunze, page 187 Here is the theorem and its "matrix analogue"  :
First of all, does the formulation of $X$'s follow from the fact that $T(X) = cX$ iff $AX = cX$, where $c$ is a characteristic value of $T$ and $A$ is the representative matrix of $T$? If so, I guess this is true only when $A$ represents $T$ relative to the standard basis of $V$ and $V=R^n$. Do I get this right?
Second, what is the relationship between the matrix $P$ and the similar matrices of $A$? I mean, what is the motivation for the construction of $P$? I think the writer's manner of telling is not explicit here.
 A: Yes, he's switching from the case of a linear transformation that might not be a matrix to a linear transformation that does operate on a set of vectors in the standard way by matrix multiplication. For some reason books use letters near the end of the alphabet (often $S$, $T$) for the former and letters near the beginning of the alphabet (often $A$, $B$) for the latter.  So the "column matrices" (most people would call these "column vectors" or just "vectors") $X$ for which $(T - c_i I)X = 0$ in the theorem become, after taking $T = A$, vectors $X$ for which $(A - c_i I) X = 0$ which as you note is the same condition as $AX = c_i X$.  (Separately, what he calls "characteristic values" are more commonly called "eigenvalues" these days, and the nonzero vectors $X$ for which $AX = c_iX$ are called "eigenvectors."  I will do this below, but it's exactly the same notion as what he's talking about in different terminology.)
Regarding the second part of the question, it really comes down to the relationship between matrices as formal objects and their properties as linear transformations.  When acting on the standard basis, any matrix $P$ can be thought of as a linear transformation sending the $i$th standard basis vector $e_i$ to the $i$th column of $P$ (call it $v_i$).  To repeat the essential point, the statements that "the $i$th column of $P$ is the vector $v_i$" and "$Pe_i = v_i$" are equivalent.  This is the heart of the matrix-theoretic aspect of the problem.  It might be the definition of matrix multiplication, or it might be a consequence of it, depending upon how your book has set this up.
So now if $v_i$, the $i$th column of $P$, was chosen to be an eigenvector of $A$ corresponding to the eigenvalue $c_i$ (as was done when the author constructed $P$ by making eigenvectors of $A$ for its columns) then you know what $A$ does to $v_i$: we have $A v_i = c_i v_i$.  And if $P$ is invertible, what would $P^{-1}$ do to that vector?  From $P e_i = v_i$ you can deduce $e_i = P^{-1} v_i$, so you know that $P^{-1}(A v_i) = P^{-1}(c_i v_i) = c_i P^{-1} v_i = c_i e_i$.
So the composite linear transformation $P^{-1} A P$ sends $e_i$ to $v_i$ (via $P$), then $v_i$ to $c_i v_i$ (via $A$), and then $c_i v_i$ to $c_i e_i$ (via $P^{-1}$).  And there's only one matrix that sends $e_i$ to $c_i e_i$ for all $i$, namely, the diagonal matrix with the list of numbers $c_i$ down the diagonal.
There's a separate question: if you weren't into matrix diagonalization in the first place, how would it occur to you to construct the matrix $P$ and even think to perform those calculations about $AP$ or $P^{-1} AP$.  I don't have a good answer, except that once you sit down long enough with the interaction between function composition and matrix multiplication, it seems like a very good or even natural idea.  And at one time it was not evident to anybody that it would be helpful to think in terms of matrix algebra at all.  We can benefit from the fact that a lot of this has been solved by others, and we can just come in later and understand what they've done. It really does fit together.
