Characteristic polynomial for the rank-$1$ perturbation

Question

Define $A=\operatorname{diag}\{a_i\}$. Introduce the polynomials $$ p(\lambda)=\det(\lambda I-A)=\prod_{i=1}^n(\lambda-a_i),\qquad p_i(\lambda)=\frac{p(\lambda)}{\lambda-a_i}=\prod_{j\ne i}(\lambda-a_j), $$ and calculate the characteristic polynomial for the rank-one perturbation $A+bk^T$ using Sylvester's determinant theorem \begin{align} \det(\lambda I-A-bk^T)=p(\lambda)(1-k^T(\lambda I-A)^{-1}b)=p(\lambda)-\sum_{i=1}^n k_ib_ip_i(\lambda). \end{align}

Can somebody please explain how did we get the last expression in the last row?

Answer 1

Since $$ p(\lambda)(\lambda I-A)^{-1} =p(\lambda)\operatorname{diag}\left(\frac{1}{\lambda-a_1},\,\frac{1}{\lambda-a_2},\ldots,\frac{1}{\lambda-a_n}\right) =\operatorname{diag}\left(p_1(\lambda),\,p_2(\lambda),\ldots,p_n(\lambda)\right), $$ we have $$ p(\lambda)k^T(\lambda I-A)^{-1}b =k^T\operatorname{diag}\left(p_1(\lambda),\,p_2(\lambda),\ldots,p_n(\lambda)\right)b =\sum_ik_ib_ip_i(\lambda). $$