Eigenvalues of $A+v_1d^T$ where $Av_1 = \lambda_1 v_1$ (shift of first eigenvalue)

Question

I have trouble to solve the following problem:

Let $A\in \mathbf{R}^{n\times n}$, $\lambda_1,\ldots,\lambda_n$ are eigenvalues of $A$, and $A v_1=\lambda_1v_1$. Let $d\in \mathbf{R}^n$, then the eigenvalues of $A+v_1d^T$ are $\lambda_1+d^Tv_1, \lambda_2,\ldots,\lambda_n$. Note: $A$ may not be diagonalizable.

I know the following facts:

1. $v_1d^T$ is of rank $1$.
2. eigenvalues of $v_1d^T$ are $v_1^Td, 0,\ldots, 0$.

But I still have no idea to prove this theorem. Please help me, thanks!

Answer 1

Sketch: Use Jordan decomposition. WLOG, assume \begin{align} A = UJU^{-1} \end{align} where \begin{align} J = \begin{pmatrix} \lambda& 1 & 0 & \dots & \dots & 0\\ 0 & \lambda & 1 & 0 & \dots & \vdots\\ \vdots & 0 & \ddots & \ddots & \dots &\vdots\\ \vdots & \vdots & \ddots& \ddots & 1 & \vdots\\ \vdots & \dots& \dots & 0 & \lambda & 1\\ 0 & \dots & \dots & \dots & 0 & \lambda \end{pmatrix} \end{align} Since $Ue_1 = v_1$ then we see that \begin{align} A+v_1d^T =&\ U(J+U^{-1}v_1d^TU)U^{-1}\\ =&\ U(J+e_1(U^Td)^T)U^{-1}. \end{align} Note that \begin{align} e_1(U^Td)^T = \begin{pmatrix} 1 \\ 0 \\ \vdots\\ 0 \end{pmatrix} \begin{pmatrix} v_1^Td & * &\dots & * \end{pmatrix} = \begin{pmatrix} v_1^Td & * &\dots & *\\ 0 & 0 & \dots & 0\\ \vdots & \vdots & \dots & \vdots\\ 0 & 0 & \dots & 0 \end{pmatrix}. \end{align}