Doubts about a question I asked a long time ago (eigenvalues) Here I posted a question about the eigenvalues of the matrix $A:=vv^t$ (where $v\in\mathbb{R}^n$). 
The question was answered but I think (after some time) that I am not satisfied.
Can someone please expand the answer? I don't understand why $A$ has rank at most $1$ and why this fact implies that $\lambda=\sum x_i^2$ is the unique eigenvalue. In addition, can I conclude that $A$ is diagonalizable?

Note that $$(vv^T)v = v(v^Tv) = \|v\|^2 v$$
so $v$ is an eigenvector with the eigenvalue $\|v\|^2 = \sum_{i=1}^n x_i^2$.
Also, explicitly
$$vv^T = \begin{pmatrix} x_1^2 & x_1x_2 & \ldots & x_1x_n \\
x_2x_1 & x_2^2 & \ldots & x_2x_n\\
\vdots & \vdots & \ddots & \vdots \\
x_nx_1 & x_nx_2 & \ldots & x_n^2\end{pmatrix}$$
so every column is a multiple of $v$, therefore the rank of $vv^T$ is at most $1$.
