advanced solutions for an elementary problem! Let $A\in M_3(\mathbb{R})$ be a matrix of rank one. Suppose that the first row of $A$ is an eigen vector of $A$. I want to show that $A$ is symmetric. 
My attemp: 
Actually its simple, for example take the first row $v = (a,b,c)$ then the second and third row would be $\alpha v$ and $\beta v$. Now put them in $Av^t = \lambda v^t$ and continue. But I am looking for other solutions or ideas. 

The issue is connected with a generic result :
The general structure of rank 1 matrices that are symmetric is $A=\pm uu^t$ for any non null (column) vector $u \in \mathbb{R^3}$.
Proof: A rank-one matrix has the following structure $A=uv^t$, where $u$ and $v$  are nonzero (column) vectors. As a consequence, $Au=(uv^t)u=\lambda u$, (showing that $u$ is naturally an eigenvector of $A$, and all eigenvectors are proportional to it) where $\lambda$ is the dot product of $u$ and $v$. 
If you impose $A^t=A$, $v$ will, for the same reason, be an eigenvector. 
Thus $u=\alpha v$ for some $\alpha$. In this way, the general solution will be $A=\alpha uu^t$, or by "incorporating"  $\alpha$ into $u$ (and taking into account its sign) $A=\pm uu^t$, as desired. 
