Is the cone of rank-$1$ matrices convex and closed? Define the following cone:
$$M = \{ x x^T : x \in \mathbb{R}^n \}$$
Is this cone convex and closed? How to prove?
 A: $M$ is precisely the set of rank one, symmetric nonnegative real matrices (EDIT and the $0$ matrix). Every nonneative symmetric real matrix is a linear combination of symmetric, nonnegative real matrices of rank one, this follows for example from the spectral theorem, so no, $M$ is not convex. Its convex hull is the convex cone of nonnegative symmetric matrices.
$M$ is closed. If $m_n=x_nx_n^T$ converges to a matrix $m$, then $m$ is obviously symmetric, and has rank $\leq 1$. Indeed, if it were of rank $>1$ there'd be two vectors $x,y$ with $(mx,my)$ linearly independent, and for $n$ great enough, $(m_nx,m_ny)$ would also have to be linearly independent, which contradicts their rank being $\leq 1$. Also it is nonnegative, since for every vector $x$, $\langle mx| x\rangle=\lim_{n\to\infty}\langle mx| x\rangle$ is a limit of nonnegative reals, so is nonnegative.
So $m$ is symmetric, nonnegative of rank $\leq 1$. If $m=0$ then $m=00^T\in M$, otherwise, $m=\lambda xx^T=yy^T$ where $\lambda>0$ is $m$'s only nonzero eigenvalue, and $x$ is a unit length vector from the line $\mathrm{E}_{\lambda}(m)=\mathrm{Im}(m)$, and $y=\sqrt{\lambda}\:.x$.
