We consider $A \in \mathcal{S}_n(\mathbb{R})$ symmetric with eigenvalues $\lambda_1 \leqslant \cdots \leqslant \lambda_n$ and corresponding orthonormal basis of eigenvectors $\left(\mathbf{w}_1, \ldots, \mathbf{w}_n\right)$. We set $B = A + \mathbf{u u}^T$ with $\|\mathbf{u}\| = 1$. We denote by $\chi_A(x) = \operatorname{det}\left(x \mathbb{I}_n - A\right)$ the characteristic polynomial of $A$, and $\chi_B(x) = \operatorname{det}\left(x \mathbb{I}_n - B\right)$ that of $B$. Show that, for all $x \in \mathbb{R} \backslash \left\{\lambda_1, \ldots, \lambda_n\right\}$, we have $$\chi_B(x) = \chi_A(x)\left(1 - \sum_{k=1}^n \frac{\left\langle \mathbf{w}_k, \mathbf{u} \right\rangle^2}{x - \lambda_k}\right).$$
We consider $A \in \mathcal{S}_n(\mathbb{R})$ symmetric with eigenvalues $\lambda_1 \leqslant \cdots \leqslant \lambda_n$ and corresponding orthonormal basis of eigenvectors $\left(\mathbf{w}_1, \ldots, \mathbf{w}_n\right)$. We set $B = A + \mathbf{u u}^T$ with $\|\mathbf{u}\| = 1$. We denote by $\chi_A(x) = \operatorname{det}\left(x \mathbb{I}_n - A\right)$ the characteristic polynomial of $A$, and $\chi_B(x) = \operatorname{det}\left(x \mathbb{I}_n - B\right)$ that of $B$. Show that, for all $x \in \mathbb{R} \backslash \left\{\lambda_1, \ldots, \lambda_n\right\}$, we have
$$\chi_B(x) = \chi_A(x)\left(1 - \sum_{k=1}^n \frac{\left\langle \mathbf{w}_k, \mathbf{u} \right\rangle^2}{x - \lambda_k}\right).$$