Let $x\neq 0$ is the eigenvector of $n\times n$ complex matrix $A$, with the eigenvalue $a$. Let $y\in \Bbb C^n$. What is the relationship between eigenvalue of $A$ and that of $A+xy'$, where $y'$ is the transpose of $y$.
By example $A=diag(1,\cdots,n)$, it seems that the eigenvalue of $A+xy'$ is $a+x'y, b$, where $\mu$ is the eigenvalue of $A$ distinct to $a$.
Let's consider the case of $A=\alpha I$, that is $A$ is a constant multiple of $I$, the identity matrix, then : \begin{align*} \operatorname{det}(A-\lambda I)&=\operatorname{det}(A+xy^{\intercal}-\lambda I) \\ &=\operatorname{det}(\alpha I+xy^{\intercal}-\lambda I)\\ &=\operatorname{det}(I(\alpha-\lambda)+xy^{\intercal})) \end{align*} Note that if $\lambda=\alpha$, then $\operatorname{det}(I(\alpha-\lambda)+xy^{\intercal}))=\operatorname{det}(xy^{\intercal}))=0$. This is true since $xy^{\intercal}$ is a rank-1 matrix. Therefore, $\lambda=\alpha$ is an eigenvalue of $\alpha I$.
Otherwise, if $\lambda\neq \alpha$, what should be the eigenvalue of $\alpha I+xy^{\intercal}$? How are they related to the eigenvalue of $\alpha I$?. Now that you determine these relations you should easily establish the result for any $n\times n$ matrix $A$
