If $A$ is idempotent and symmetric, then $A=BB^T$ where $B^TB=I$.
I know that since $A$ is symmetric, then it is positive semidefinite, i.e.
$x^TAx\ge 0 \forall x$,
then there is a $B$ such that $A=B^TB$.
But isn't this true only if $A$ is of full rank?
Best Answer
That is not true.
$A=0$ is idempotent, as $0^2=0$. If $A=BB^T$ and $B^TB=I$, then $$I=B^TBB^TB=B^T(BB^T)B=B^TAB=B^T0B=0.$$
Try also, $$ A=\left(\begin{matrix} 1&0\\0&0\end{matrix}\right). $$ Then $A^2=A$. If $A=BB^T$ and $B^TB=I$, then $$I=B^TBB^TB=B^T(BB^T)B=B^TAB,$$ which implies that $A$ is nonsingular - contradiction.
However, assuming that $A$ is non-singular, then $A^2=A$ implies that $A(A-I)=0$, and in turn that $$ A-I=A^{-1}A(A-I)=A^{-1}0=0, $$ and hence that $A=I$.