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?
 A: A is diagonizable because it is real valued and symmetric $A^T=A$ (This is the spectral theorem)
$A=vv^T$ so $A^T=(vv^T)^T=(v^T)^Tv^T=vv^T=A$
https://en.wikipedia.org/wiki/Spectral_theorem
A: To see that $v v^T$ has rank at most one, we see that for any arbitrary vector $u$ ,we have:
$$
(vv^T)x = v(v^Tx)
$$
The right hand side is a scalar times $v$. So every vector $x$ is mapped to a scalar multiple of $v$, with the scalar determined by $x$, meaning that the linear transformation represented by $vv^T$ has rank $1$ and so $vv^T$ has rank $1$.
A: 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 $1$.
A: $A$ has rank at most $1$ because all rows are multiples of $(x_1,\dots,x_n)$.
Therefore, the kernel of $A$ has dimension at least $n-1$, leaving room for at most one other eigenvalue.
$A$ is diagonalizable by taking a basis of the kernel of $A$ and adding an eigenvector for $\lambda$.
A: You want to write $v$ as a column vector. Do the computation for any vector you like, and you will notice that the columns of $A$ are just a multiple of $v$.
A: Rank $1$: For any vector $a$, note that $v^Ta$ is a scalar, so $vv^Ta$ is a scalar multiple of $v$. So the range of $vv^T$ is within the span of $v$, which is one-dimensional, so $vv^T$ has rank $1$.
All but $1$ eigenvalue is $0$: Since the rank of $vv^T$ is $1$, the nullspace has dimension $n-1$, by the rank-nullity theorem. That means we can find $n-1$ linearly independent vectors in the nullspace. Since every vector in the nullspace has eigenvalue $0$, $0$ is an eigenvalue with multiplicity $n-1$. That leaves room for only one more eigenvalue, which we have already shown is $\sum x_i^2$.
Diagonalizability: As we have shown above, we can find $n-1$ linearly independent eigenvectors of $0$. We can also find one eigenvector of $\sum x_i^2$, to make a basis in which $vv^T$ is diagonal.
