Vrftjkry

I have trouble to solve the following problem:

Let $Ain mathbf{R}^{ntimes n}$, $lambda_1,ldots,lambda_n$ are eigenvalues of $A$, and $A v_1=lambda_1v_1$. Let $din mathbf{R}^n$, 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!

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}