Express $\operatorname{trace}(B'XB)$ in terms of $A$ and $B$
Andrew Henderson 
Given $A\in\Bbb R^{n\times n}$, $B\in\Bbb R^{n\times m}$, and $X>0$, s. t. $X=A'XA-A'XB(I+B'XB)^{-1}B'XA,$ where $A'$ is $A$ transpose.
Is it possible to express $\operatorname{trace}(B'XB)$ in terms of $A$ and $B$ only (without $X$)?
If it helps, $(A,B)$ is stabilizable. Even for diagonal $A$, the answer is not obvious.
My attempt:
I only have few equalities that I managed to deduce:
- $\operatorname{trace}(B'XB)=\operatorname{trace}(AX^{-1}A'X)-\operatorname{trace}(I)=\operatorname{trace}(AX^{-1}A'X)-n.$
- $\operatorname{trace}(B'XB)=\sum\limits_{i=1}^m(B_i'XB_i)$, where $B_i$ is the $i$'th column of $B$.
Let $A=\begin{bmatrix}a_1&&\\&a_2&\\ &&a_2\end{bmatrix}$, then $\operatorname{trace}(B'XB)=a_1^2a_2^2 + a_2^2 -2$. (i.e. independent of $B$)
$\det(A_1)^2+\cdots+\det(A_m)^2\geqslant \operatorname{trace}(AX^{-1}A'X)\geqslant m\sqrt[m]{\det(A)^2}$. To prove this part, we can do Wonham decomposition on $(A,B)$ then use 1 and 2 together with geometric mean.
Is there any tighter bound than 4?
$\endgroup$ 61 Answer
$\begingroup$From $X = A^\mathsf{T}XA - A^\mathsf{T}XB(I + B^\mathsf{T}XB)^{-1}B^\mathsf{T}X A$, we have$X = A^\mathsf{T}(I + XBB^\mathsf{T})^{-1}XA$, and $I + XBB^\mathsf{T} = XAX^{-1}A^\mathsf{T}$, and $X^{-1} + BB^\mathsf{T} = AX^{-1}A^\mathsf{T}$. Let $Y = X^{-1}$. We have$$BB^\mathsf{T} = AYA^\mathsf{T} - Y \tag{1}$$which is written as$\mathrm{vec}(BB^\mathsf{T}) = (A \otimes A - I)\mathrm{vec}(Y)$ where $\otimes$ denote the Kronecker product. See:
Since the eigenvalues of $A$ are larger than $1$, we know that zero is not an eigenvalues of $(A \otimes A - I)$and thus $(A \otimes A - I)$ is non-singular. Thus, we have $\mathrm{vec}(Y) = (A \otimes A - I)^{-1}\mathrm{vec}(BB^\mathsf{T})$, from which, we get $Y$. We have $\mathrm{Tr}(B^\mathsf{T}XB) = \mathrm{Tr}(B^\mathsf{T}Y^{-1}B)$.
$\endgroup$ 3
