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 are the eigenvalues of $A+v_1d^T$ equal to $\lambda_1+d^Tv_1,\lambda_2,\ldots,\lambda_n$ ?
This was my previous answer, it was wrong. See the correct statement and proof here
Hint: This is only generally true if we also know that $v_i$ is perpendicular to $d$ for $i=2,3,\dots,n$. You should be able to construct a $2 \times 2$ counterexample.