Matrix decomposition the other way First of all, this is no useful way to decompose a matrix -
you need to know the eigenvalues beforehand. But it popped up
naturally during my knot theory dabblings.
Assume that you know the characteristic equation
$$\prod_{i=1}^n (S - e_i I) = 0$$
with $S$ being an $n \times n$ matrix, $I$ the $n \times n$ identity matrix and $e_1, e_2, \dotsc$ the eigenvalues of $S$.
Define the matrices
$$T_i = \prod_{j \neq i} \frac{1}{e_i-e_j} (S - e_j I)$$
Now $T_i T_j=0$ if $i \neq j$ (obvious) and $T_i.T_i=T_i$ (needs proof),
to the effect that you can compute
$$S^k = \sum_{i=1}^n e_i^k T_i$$
for any $k$ (obvious again).     
Question: Under what conditions does this scheme work? Equal eigenvalues are no problem, this just reduces the number of $T_i$ needed, but I've got a hunch that equal eigenvalues which are defective (off-diagonal elements in the Jordan decomposition) will be ruinous. (An elegant proof of $T_i.T_i=T_i$ is also welcome, but on gunpoint I'll probably come up with one myself :-)
 A: The decomposition should work if and only if the minimal polynomial of the matrix can be factored into pairwise non-proportional linear factors.
Let me tell you how algebraists think about this (depending on your background, you might find everything here trivial): You have a matrix $S\in\mathrm{M}_n\left(k\right)$, where $k$ is a field. Let $m\in k\left[X\right]$ be the minimal polynomial of $S$. Then, the $k$-algebra $k\left[S\right]$ (this is the $k$-subalgebra of $\mathrm{M}_n\left(k\right)$ generated by $S$) is isomorphic to the $k$-algebra $k\left[X\right] / \left(m\right)$. (Here, $\left(m\right)$ denotes the ideal of $k\left[X\right]$ generated by $m$. As much as I dislike this notation, it is short.)
Now assume that $m$ can be factored into pairwise non-proportional linear factors, i. e. that we have $m=\lambda p_1p_2...p_u$ for some $\lambda\in k$ and some pairwise non-proportional linear polynomials $p_1,p_2,...,p_u$. Then,
$k\left[X\right] / \left(m\right) = k\left[X\right] / \left(p_1p_2...p_u\right)$
$\cong \left(k\left[X\right] / \left(p_1\right)\right) \times \left(k\left[X\right] / \left(p_2\right)\right) \times ... \times \left(k\left[X\right] / \left(p_u\right)\right)$
(where $\times$ means the direct product of $k$-algebras) by the Chinese Remainder Theorem for $k$-algebras. Each $k\left[X\right] / \left(p_i\right)$ is isomorphic to $k$ (because $p_i$ is linear), so that you obtain
$k\left[X\right] / \left(m\right) \cong k \times k \times ... \times k$ (with $u$ times $k$).
Together with $k\left[S\right] \cong k\left[X\right] / \left(m\right)$, this leads to
$k\left[S\right] \cong k \times k \times ... \times k$.
Now, for every $i$, the element $\left(0,0,...,0,1,0,0,...,0\right)$ (with $1$ on the $i$'th place, and $0$ on every other place) of $k \times k \times ... \times k$ corresponds to your $T_i\in k\left[S\right]$ under this isomorphism. Your construction of $T_i$ is pretty much equivalent to the standard constructive proof of the Chinese Remainder Theorem.
On the other hand, if $m$ cannot be factored into pairwise non-proportional linear factors, then $k\left[S\right]$ is not isomorphic to $k \times k \times ... \times k$. However, if $m$ can be factored into linear factors (for example, this happens if $k$ is algebraically closed), then at least it is isomorphic to a direct product of $k$-algebras isomorphic to $k\left[X\right] / \left(X^d\right)$ for various $d$ (each of these $k$-algebras corresponds to a Jordan block of $S$, so you still have elements like $\left(0,0,...,0,1,0,0,...,0\right)$, but they should not be as simple as your $T_i$ anymore, and they do not linearly span that direct product, so you shouldn't expect a formula as simple as $S^k = \sum_i T_i\cdot\left( \text{some constant}\right)^k$ to hold.
A: I have just seen Darij Grinberg's answer appear as I was writing, but let me give a slightly different (but closely related) perspective.
I find your text difficult to read-(note added later:before real latex inserted)-but I think you have rediscovered a standard way to diagonalise a semisimple matrix. You are implicitly assuming that $S$ has distinct eigenvalues. It's also related to how to invert a Van der Monde Matrix, and to Lagrange interpolation.
 I like to think of it this way. Let $A$ be a cyclic finite dimensional algebra over a field $F$,
generated by an element $T$, which has minimum polynomial $p(x) = \prod_{i=1}^{n}(x-\lambda_{i})$,
where the $\lambda_i$ are distinct elements of $F$. Hence $A$ has dimension $n$, and has an
$F$-basis $\{I,T,\ldots,T^{n-1} \}.$ There are $n$ different non-zero algebra homomorphisms 
from $A$ to $F$, say $\{\mu_i : 1 \leq i \leq n \}$, where $T\mu_i = \lambda_i$ 
and $I\mu_i = 1$ for each $i$.
This gives an algebra homomorphism from $A$ to $F \times F \ldots \times F$ (n copies), which is an isomorphism by dimension. Hence the algebra $A$ is commutative semisimple, since any nilpotent element is sent to zero by each $\mu_i$, so must be zero. Furthermore, for each $i$,
the element $$E_i = \prod_{j \neq i} \frac{T - \lambda_{j}I}{\lambda_{i} - \lambda_{j}}$$
clearly has` $E_{i}\mu_{j} = \delta_{ij}$ for $1 \leq j \leq n.$ Hence $E_{i}^{2}-E_{i}$
is annihilated by each $\mu_{j}$, so is zero, and each $E_{i}$ is idempotent.
The connection with inverting a Van der Monde matrix is as follows: evaluating the linear 
characters at powers of $T$ shows that for $0 \leq i \leq n-1$, we have:
$T^{i} = \sum_{j=1}^{n} \lambda_{j}^{i} E_{j}.$. This shows that the matrix of coefficients 
to express the basis $\{T^{i}: 0 \leq i \leq n-1 \}$    in terms of the basis 
$\{E_{i} : 1 \leq i \leq n\}$ is the Van der Monde matrix associated to $\{\lambda_{1},
\ldots \lambda_{n}\}$. The matrix of coefficients needed to express the basis 
$\{E_{i} : 1 \leq i \leq n\}$ in terms of the basis $\{T^{i}: 0 \leq i \leq n-1 \}$
is therefore the inverse of that Van der Monde matrix. But this matrix can be easily read
from the expressions 
$$E_i = \prod_{j \neq i} \frac{T - \lambda_{j}I}{\lambda_{i} - \lambda_{j}}$$
for $1 \leq i \leq n$.
Your argument (as it stands) will fail if the matrix $S$ has a minimum polynomial which is
not multiplicity free. This corresponds the the fact that the cyclic $F$-algebra $A$,
generated by an element $T$ with the same minimum polynomial as $S$ (assuming $F$ contains
all roots of this polynomial) is no longer a commutative semi-simple algebra. This can be
seen directly, since if the distinct roots of the minimum polynomial are 
$\lambda_1,\ldots \lambda_m$, then $\prod_{i=1}^{m} (T-\lambda_{i}I)$ is non-zero
by hypothesis, but is clearly nilpotent.
A: This is a fairly standard result in the theory of diagonalizable linear operators, sometimes known as the spectral decomposition theorem for diagonalizable operators. Indeed, a linear operator over some field is diagonalizable if and only if it has a decomposition of this form. You can read about this in Hoffman & Kunze's "Linear Algebra" (specifically, the chapter "invariant sum decompositions"). The following theorem (quoted from the book) sums most of it up:

Let $T$ be a linear operator on a
  finite-dimensional space $V$. If $T$
  is diagonalizable and if
  $c_1,\dots,c_k$ are the distinct
  eigenvalues of $T$, then there exist
  linear operators $E_1,\dots,E_k$ on
  $V$ such that:
  
  
*
  
*$T = c_1 E_1 + \cdots + c_k E_k$
  
*$I = E_1 + \cdots + E_k$
  
*$E_i E_j = 0$ if $i \ne j$.
  
*$E_i ^2 = E_i$ for all $i$.
  
*The image of $E_i$ is the space of $T$-eigenvectors with eigenvalue
  $c_i$.
  
  
  Conversely,if there exist $k$ distinct
  scalars $c_1,\dots,c_k$ and $k$
  non-zero linear operators
  $E_1,\dots,E_k$ which satisfy
  conditions 1,2,3, then $T$ is
  diagonalizable, $c_1,\dots,c_k$ are
  the distinct eigenvalues of $T$ and
  conditions 4,5 are satisfied also.

Note that given such a decomposition for $T$, we have $f(T) = \sum f(c_i) E_i$ for any polynomial $f$ over the field. Fix some $1 \le i \le k$ and consider the polynomial $f (x) = \prod_{j \ne i} \frac{x - c_j}{c_i - c_j}$. This polynomial satisfies $f(c_r) = \delta_{r,i}$ and thus $f(T) = E_i$. So the projections $E_i$ are necessarily of the form you found.
By standard theory, $T$ is diagonalizable if and only if its minimal polynomial factors into distinct linear factors. Now, one may ask what kind of decomposition we can get if the minimal polynomial of $T$ factors into linear factors, not necessarily distinct. In this case, one can still form the sum $c_1 E_1 + \cdots + c_k E_k$ (where the $E_i$ are defined either by your formula or by property 5 above), but it won't be equal to $T$. Rather, it will differ from $T$ by a nilpotent operator (an operator $N$ with $N^m = 0$  for some integer $m$). More generally, a linear operator on a fin. dimensional vector space, over an algebraically closed field, can be written uniquely as the sum of a diagonalizable operator (called its diagonalizable part) and a nilpotent operator (called its nilpotent part) which commute with with one another. In this case $c_1 E_1 + \cdots + c_k E_k$ gives the diagonalizable part of $T$.
If the field is not algebraically closed then one may not be able to form the sum $c_1 E_1 + \cdots + c_k E_k$ at all, since some of the eigenvalues of $T$ may not be in the field. Nevertheless, there is still an analogous decomposition for $T$ in this case (at least when the field has characteristic zero, I guess), which represents $T$ (uniquely) as the sum of a "semisimple" operator and a nilpotent operator which commute with one another. Here "semisimple" is a property of operators which is equivalent to being diagonalizable if the field is algebraically closed, but is otherwise more involved.
