TITLE: Are all rank-1 positive semidefinite matrices the result of an outer product $vv^T$?
QUESTION [9 upvotes]: If $v$ is a column vector in $\mathbb{R}^n$, then it is obvious that $V = vv^T$ is a rank-1 positive semidefinite matrix.  I am wondering if the converse is true: if $V$ is a square real rank-1 positive semidefinite matrix, can it be written in the form $vv^T$?

REPLY [13 votes]: Hint:  consider an eigenvector ...

REPLY [10 votes]: All rank one matrices are $w v^T,$ both nonzero and not required equal; if the vectors have the same number of entries the matrix is square. If square and symmetric, $w= \lambda v$ for a nonzero real number $\lambda.$ As in Robert's answer, if $\lambda > 0$ you get positive semidefinite, if $\lambda < 0$ you get negative semidefinite. In either case you can take $x = v \sqrt {|\lambda|}$ with the result that either $w v^T = x x^T$ or $w v^T = - x x^T.$