We now consider the symmetric matrix $A$. By virtue of the spectral theorem, we denote by $\lambda_1 \leqslant \cdots \leqslant \lambda_n$ the eigenvalues of $A$, and $\left(\mathbf{w}_1, \ldots, \mathbf{w}_n\right)$ a corresponding orthonormal basis of eigenvectors. (a) Show that $$A = \sum_{k=1}^n \lambda_k \mathbf{w}_k \mathbf{w}_k^T$$ (b) Show that for all $x \in \mathbb{R} \backslash \left\{\lambda_1, \ldots, \lambda_n\right\}$, we have $$\left(x \mathbb{I}_n - A\right)^{-1} = \sum_{k=1}^n \frac{1}{x - \lambda_k} \mathbf{w}_k \mathbf{w}_k^T$$
We now consider the symmetric matrix $A$. By virtue of the spectral theorem, we denote by $\lambda_1 \leqslant \cdots \leqslant \lambda_n$ the eigenvalues of $A$, and $\left(\mathbf{w}_1, \ldots, \mathbf{w}_n\right)$ a corresponding orthonormal basis of eigenvectors.\\
(a) Show that
$$A = \sum_{k=1}^n \lambda_k \mathbf{w}_k \mathbf{w}_k^T$$
(b) Show that for all $x \in \mathbb{R} \backslash \left\{\lambda_1, \ldots, \lambda_n\right\}$, we have
$$\left(x \mathbb{I}_n - A\right)^{-1} = \sum_{k=1}^n \frac{1}{x - \lambda_k} \mathbf{w}_k \mathbf{w}_k^T$$