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, 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.

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.

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, 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 $I\lambda_i = 1$ 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.

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, 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.