What is the solution to Discrete Algebraic Riccati Equation (DARE) when $A$ is identical? Let $X$, $R$ and $Q$ be $n$-by-$n$ positive definite matrices, and $B$ is an $m$-by-$n$ matrix. Then is there any closed form solution to the following DARE w.r.t. $X$?
$$XB^T (BXB^T + R)^{-1}BX = Q.$$
Thanks!
PS: This equation is a special case of a standard DARE
$$X = AXA^T - AXB^T (BXB^T + R)^{-1}BXA^T + Q,$$
with $A = I$.
 A: Note that $R\in S_{m,m}$. Let $(1)$: $XB^T(BXB^T+R)^{-1}BX=Q$.
EDIT. I correct several mistakes. Note that $B^T(BXB^T+R)^{-1}B$ is invertible, and, consequently, necessarily $n\leq m$ and $rank(B)=n$; moreover $E\in M_n,BEB^T=0$ implies $E=0$.
Thus the equation $(1)$ in the unknown $X$ is equivalent to $(*)$ $BXB^T(BXB^T+R)^{-1}BXB^T=BQB^T$.
We consider the new unknown $Y=BXB^T\in S_{m,m}$ and let $U=BQB^T\in S_{m,m}$. Unfortunately $Y,U\geq 0$ but, if $n<m$, then they are not invertible ($rank(Y)=rank(U)=n$). Remark that $Y+R>0$.
Thus $(*) \Leftrightarrow Y(Y+R)^{-1}Y=U$. When $n=m$, this last equation reduces easily to a standard Riccati equation. 
The case  $n<m$. One has $(Y+R)^{-1}=(I+R^{-1}Y)^{-1}R^{-1}$; finally 
$(*) \Leftrightarrow R^{-1}Y(I+R^{-1}Y)^{-1}R^{-1}Y=R^{-1}U$. We put $Z=R^{-1}Y$ and deduce that
$(*) \Leftrightarrow Z(I+Z)^{-1}Z=R^{-1}U \Leftrightarrow Z^2-(R^{-1}U)Z-R^{-1}U=0$, that is a Riccati equation.
Standard softwares give an approximation of $Z=R^{-1}Y$ (of course, there is no closed form for a solution $Z$) and we deduce an approximation of $Y$; since $B^T$ is onto  and $B$ is one to one, the equation in $X$: $Y=BXB^T$ has a unique solution which is easy to find. 
