If $A$ is a square $n\times n$ matrix, with $\lambda_1,\ldots,\lambda_n$ being the eigenvalues of $A$, $v_1$ being the eigenvector associated with eigenvalue $\lambda_1$, and $d$ the column vector of dimension $n$, then can anyone provide a counterexample to show the following statement is false: the eigenvalues of $A+v_1d^T$ are equal to $\lambda_1+d^Tv_1,\lambda_2,\ldots,\lambda_n$ ?
Any help or direction is appreciated, thank you!
There are no counterexamples, this statement is apparently true (assuming that $u_1$ is a unit eigenvector).
One quick proof uses Schur triangularization: given $A$, find a change of basis $U$ such that $U^*AU$ is upper-triangular. In particular, $$ U^*AU = \pmatrix{\lambda_1 && *\\ &\ddots\\&&\lambda_n} $$ Then, we find that $$ U^*(A + u_1d^*)U = U^*AU + (U^*u_1)(U^*d)^* = U^*AU + e_1(U^*d)^* $$ Let $w = U^*d$, we then have $$ U^*(A + u_1d^*)U= \pmatrix{\lambda_1 + w_1 &&& *\\&\lambda_2 \\&&\ddots\\&&&\lambda_n} $$ The statement follows.
Note: we have $w_1 = d^Tv_1$ since $$ w_1 = e_1^*w = (U^*v_1)^*(U^*d) = v_1^* d = v_1^Td $$