as $A$ is hermitian positive definite so all eigen values are real and strictly positive. we can diagonalize $A=PDP^{-1}$ for some invertible $P$ and we can chose $B=P\sqrt{D}P^{-1}$ and hence $a$ is true?
$BB'$ is symmetric that is true, I am not sure about positive definiteness, let $X$ be any eigen vector corresponding to the eigen value $\lambda$ then $BB^TX=\lambda X\Rightarrow XBB^TX=\lambda XX^T\Rightarrow X(BB^T-\lambda I)X^T=0$.well please help.