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.

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@ Arturo Thanks.... – Beat Dec 13 '11 at 20:57
So...? You need to find the eigenvalues of $A$, the Frobenius covariants of $A$, and then apply the formula $$f(A)=f(\lambda_1)A_1 + f(\lambda_2)A_2$$ with $f(t)=t^{100}$. – Arturo Magidin Dec 13 '11 at 20:57
What do you mean by Sylvester's formula? Alternatively, since $A$ is symmetric you can diagonalize it and find $A^{100}$ easily. – Manos Dec 13 '11 at 20:57
@ArturoMagidin: The Frobenius covariants are the matrices $uu^T$ where $u$ is eigenvector? – Manos Dec 13 '11 at 20:59
@Manos: Here's Wikipedia's entry. Here, the Frobenius covariant, if the eigenvalues $\lambda_1$ and $\lambda_2$ are different, are $$\frac{1}{\lambda_2-\lambda_1}(A-\lambda_1 I)\text{ and }\frac{1}{\lambda_1-\lambda_2}(A-\lambda_2 I).$$ – Arturo Magidin Dec 13 '11 at 21:02

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).

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@ Arturo, you are right. I had the same. Thanks – Beat Dec 14 '11 at 11:44