Eigenvector and eigenvalue of $\mathbf A+\lambda\mathbf I$ where $\mathbf A=\mathbf{vv}^\top$ 
Given $\mathbf v=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}$ and $\mathbf A=\mathbf{vv}^\top$, find the eigenvectors and eigenvalues of $\mathbf A+\lambda\mathbf I$.

My current work progress is:
Since $(\mathbf A+\lambda\mathbf I)\mathbf v =\mathbf{Av}+\lambda\mathbf v=\mathbf v(\mathbf v^\top\mathbf v) + \lambda\mathbf v=\mathbf v \|\mathbf v\|+ \lambda\mathbf v=(\|\mathbf v\|+\lambda)\mathbf v$ ,
so one of the eigenvalue is $\lambda_1=(\|\mathbf v\|+\lambda)$ with corresponding eigenvector $\mathbf v_1=\mathbf v$.
I'd like to find out other pairs of eigenvalues and eigenvectors. I first calculate the trace.
$$\mathrm{Tr}\,(\mathbf A)=n\lambda+\|\mathbf v\|=\lambda_1+(n-1)\lambda$$
Hence, the rest of $n-1$ eigenvalues has sum $(n-1)\lambda$. 
From here, I do not know how to proceed.


*

*Can I assume the rest of eigenvalues are $\lambda$ with multiplicities $n-1$? Why?

*How to find corresponding eigenvectors?


[Note1] This is like a follow up question regarding this one.
[Note2] Found later on there is a question but people focus on linking it to Note1.

Let $\mathbf{v}^{\perp}$ be the subspace of $\mathbb{R}^n$ of all vectors orthogonal to $\mathbf{v}$.  (The proof that $\mathbf{v}^{\perp}$ is indeed a subspace is often proven or given as homework in most undergraduate linear algebra textbooks).
Let $\mathbf{w}_1, ..., \mathbf{w}_{n-1}$ be a basis of $\mathbf{v}^{\perp}$.  Then by definition of orthogonal, $\mathbf{v}^T\mathbb{w_i} = 0$ for all $i$, so
$$ (A + \lambda I)\mathbf{w}_i = \mathbf{v}(\mathbf{v}^T\mathbf{w}_i) + \lambda \mathbf{w}_i = \lambda \mathbf{w}_i $$
Thus $\{\mathbf{w}_1,...,\mathbf{w}_{n-1}\}$ are your other eigenvectors with eigenvalue $\lambda$.
