Can someone think of an easy way to prove (or disprove) the above conjecture?
For $2\times 2$ matrices, the conjecture is trivially true. For $3\times 3$ matrices, the non-negativity of the determinant can be used to show that the conjecture holds. But I don't know the answer for higher dimensional matrices. Maybe some kind of induction argument is required to prove the conjecture.
Let $A$ be $HPSD$ and $n\times n$
then $A$ has all ones on its diagonal (why?)
with eigenvalues in the usual ordering
$\lambda_1 \geq \lambda_2\geq ... \geq \lambda_n \geq 0$
$n^2$
$= \Big(1+1+...+1\Big)^2 $
$= \Big(\text{trace}\big(A\big)\Big)^2 $
$= \big(\lambda_1+ \lambda_2+...+\lambda_n\big)^2$
$\geq \lambda_1^2+ \lambda_2^2+...+\lambda_n^2 $
$= \text{trace}\big(A^2\big) $
$=\Big \Vert A\Big \Vert_F^2$
$= n^2$
this inequality is met with equality
$\implies 0=\lambda_2=...=\lambda_n$
$\implies \lambda_1 = n$
and $A$ is diagonalizable, hence it is rank 1.