Power of a matrix using Sylvester's Formula I have been thinking about this question and I'm really confused, I have gone through past solutions and I really understand those, but this, I don't understand. I'm to use Sylvester's formula to find $ A^{100}$ where the matrix is
$$A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \\ \end{pmatrix}$$
Sylvester's formula is given as $ P([A]) = \sum\limits_{k=1}^{n} P(\lambda _{k}) Z_{k}([A])$ and $$ Z_{k}([A])=\frac{\prod\limits_{r\neq k} (A - \lambda_{k}I)}{\prod\limits_{r\neq k} (\lambda_{r}- \lambda_{k}I)} , k= 1, \dots, 100$$ $ \lambda_{k}$ are the eigenvalues.

The Sylvester formula, as you write, is that if $A$ is diagonalizable, and $f(t)$ is a polynomial (in fact, any analytic function), then
$$f(A) = \sum_{i=1}^n f(\lambda_i)A_i$$
where $\lambda_1,\ldots,\lambda_n$ are the eigenvalues of $A$ (hence, $n$ is the size of the matrix), and $A_i$ is the Frobenius covariant of $A$,
$$A_i = \prod_{\stackrel{j=1}{j\neq i}}^n \frac{1}{\lambda_i-\lambda_j}(A-\lambda_jI).$$
Here, your matrix is $2\times 2$, so you will have
$$f(A) = f(\lambda_1)A_1 + f(\lambda_2)A_2,$$
where
$$A_1 = \frac{1}{\lambda_1-\lambda_2}(A-\lambda_2I),\qquad A_2 = \frac{1}{\lambda_2-\lambda_1}(A-\lambda_1I).$$
So the first step is to find the two eigenvalues; then compute the two Frobenius covariants; then apply the formula to $f(t)=t^{100}$. 
You can compute the characteristic polynomial directly (it's $(2-t)^2 - 1 = t^2-4t+3$) and find the eigenvalues; or any of the standard tricks: the determinant of $A$ is $3$, the trace is $4$, so you want to numbers that add up to $4$ and whose product is $3$: $\lambda_1=1$ and $\lambda_2 = 3$ will do.
So then
$$\begin{align*}
A_1 &= \frac{1}{1-3}(A-3I) = -\frac{1}{2}\left(\begin{array}{rr}
-1&1\\
1& -1
\end{array}\right)\\
A_2 &= \frac{1}{3-1}(A-I) = \frac{1}{2}\left(\begin{array}{cc}
1&1\\
1& 1
\end{array}\right).
\end{align*}$$
So the Sylvester formula tells you that
$$A^{100} = f(A) = \left(-\frac{f(1)}{2}\left(\begin{array}{rr}-1&1\\1&-1\end{array}\right)\right) + \left(\frac{f(3)}{2}\left(\begin{array}{cc}1&1\\1&1
\end{array}\right)\right).$$

Alternatively, $A$ is diagonalizable, with eigenvectors $(1,-1)$ (corresponding to $\lambda_1=1$) and $(1,1)$ (corresponding to $\lambda_2=3$). That means that
$$
\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right) A \left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right) = \left(\begin{array}{cc}1 & 0\\0 & 3
\end{array}\right).$$
Since $(PAP^{-1})^n = PA^nP^{-1}$, then
$$\begin{align*}
 \left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right) A^{100} \left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right) &= \left(\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}
\end{array}\right) A \left(\begin{array}{rr}
1 & 1\\
-1 & 1
\end{array}\right) = \left(\begin{array}{cc}1 & 0\\0 & 3
\end{array}\right)\right)^{100} \\
&= \left(\begin{array}{cc}
1& 0\\
0 & 3
\end{array}\right)^{100}.
\end{align*}$$
Computing the last matrix is easy, and then we just need to multiply by
$$\left(\begin{array}{rr}1 & 1\\-1 & 1
\end{array}\right)$$ on the left and $$\left(\begin{array}{rr}
\frac{1}{2} & -\frac{1}{2}\\
\frac{1}{2} & \frac{1}{2}\end{array}\right)$$
on the right to get the value of $A^{100}$. (But this does not use Sylvester's formula).
