Suppose that $A$ is a $n \times n$ matrix with $n$ different eigenvalues $\lambda_k.$ Corresponding eigenvectors are denoted as $x_k$, $x_k^Tx_k =1.$ Now $A=X\Lambda X^{-1}$. Denote $Q=X^{-1}$.
Denote $R_i = x_i \cdot q_i^T.$ Where $q_i^T$ is ith row of $Q$.
How to prove that: 1) $R_1+...+R_n=I?$
2) $(I-A)^{-1} = \displaystyle\sum_{k=1}^n \dfrac{R_k}{1-\lambda_k}?$
This decomposition might be of use: $$ \begin{align} (I-A)^{-1} = &\; (I-X\Lambda X^{-1})^{-1}\\ = &\; (X(I-\Lambda)X^{-1})^{-1} \\ = &\; X(I-\Lambda)^{-1}X^{-1} \end{align} $$
...and note that $(I-\Lambda)^{-1}$ is easily found if $\Lambda$ is diagonal.