The partial fraction expansion of a 4x4 matrix The partial fraction expansion of a matrix is given by $$(I\xi-A)^{-1}=\sum_{i=1}^{N}\sum_{j=1}^{n_{i}}T_{ij}\frac{1}{(\xi-\lambda_{i})^{j}}$$,
$T_{ij}\in\mathbb{R}^{n\times n}$, $\lambda_{i}$ the eigenvalues of the matrix $A$ and $n_{i}$ the multiplicity of the respective eigenvalues.
Take $$A=\begin{bmatrix}0 & 1 & 0 & 0 \\ -1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -4 & 0\end{bmatrix}$$
I already calculated the eigenvalues as $\lambda_{1}=-2i$, $\lambda_{2}=-i$, $\lambda_{3}=i$ and $\lambda_{4}=2i$, each with multiplicity of 1. But I am having trouble with determining the respective $T_{ij}$.
 A: In your case, the multiplicity of each eigenvalue is one so you have
$$ (I\xi-A)^{-1}=\sum_{i=1}^{N} T_{i}\frac{1}{(\xi-\lambda_{i})}. $$
The matrix $A$ is diagonalizable so you have a basis $(v_1, \ldots, v_N)$ of eigenvectors of $A$ with $Av_i = \lambda_i v_i$. Let us multiply the equation above by $v_j$:
$$ \frac{v_j}{\xi - \lambda_j} = (I\xi - A)^{-1}(v_j) = \sum_{i=1}^N \frac{T_j v_j}{\xi - \lambda_i}. $$
It is easy to see that this equation will be satisfied if $T_i v_j = \delta_{ij} v_j$. Since $(v_1, \ldots, v_n)$ is a basis, this determines $T_i$ uniquely and allows you to find them explicitly given the eigenvectors.
A: For your particular case, you have
$$(xI-A)^{-1}=\frac{T_1}{x-\lambda_1}+\frac{T_2}{x-\lambda_2}+\frac{T_3}{x-\lambda_3}+\frac{T_4}{x-\lambda_4}$$
For $x$ large, you can use power series in $1/x$:
$$\sum_{n\geq 0} \frac{A^n}{x^{n+1}}=\sum_{n\geq 0}\frac{T_1\lambda_1^n+T_2\lambda_2^n+T_3\lambda_3^n+T_4\lambda_4^n}{x^{n+1}}$$
You get $$A^n=T_1\lambda_1^n+T_2\lambda_2^n+T_3\lambda_3^n+T_4\lambda_4^n$$
Now write these relations for $n=0,1,2,3$, and you have a system with a Vandermonde determinant. 
