# Express $\operatorname{trace}(B'XB)$ in terms of $A$ and $B$

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:

1. $$\operatorname{trace}(B'XB)=\operatorname{trace}(AX^{-1}A'X)-\operatorname{trace}(I)=\operatorname{trace}(AX^{-1}A'X)-n.$$
2. $$\operatorname{trace}(B'XB)=\sum\limits_{i=1}^m(B_i'XB_i)$$, where $$B_i$$ is the $$i$$'th column of $$B$$.
3. 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 $$B$$)

4. $$\det(A_1)^2+\cdots+\det(A_m)^2\geqslant \operatorname{trace}(AX^{-1}A'X)\geqslant m\sqrt[m]{\det(A)^2}$$. 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?

• Use \operatorname{trace} and \det. Jun 11, 2020 at 7:00
• Any information about $A$? Jun 14, 2020 at 4:49
• @RiverLi We assume that $(A,B)$ is stabilizable. Also we can assume that $(A,B)$ is in wonham or cyclic decomposition form, i.e. $A$ is upper block triangular or block diagonal and $B$ is upper block triangular with size of blocks are according to size of blocks in $A$. If it also doesn't help we can work on the case when $A$ is diagonal, but some diagonal elements must repeat, if diagonal elements don't repeat I have the answer.
– Lee
Jun 14, 2020 at 7:55
• @RiverLi also eigenvalues of $A$ are larger than $1$
– Lee
Jun 14, 2020 at 7:56
• @Lee I wrote something. Jun 14, 2020 at 9:59

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: https://en.wikipedia.org/wiki/Kronecker_product
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 $$\mathrm{Tr}(B^\mathsf{T}XB) = \mathrm{Tr}(B^\mathsf{T}Y^{-1}B)$$.
• Hi, I am trying to continue this problem for specific case when A is diagonal. And simulation shows that in that case, $\mathrm{Tr}(B'XB)$ is always independent of $B$. I tried to prove it, and can see since we take inverse of $Y$, all $B$ somehow cancel each other, but I could prove it. Can you please help me?
• @Lee It is interesting if $\mathrm{Tr}(B^\mathsf{T}XB)$ is independent of $B$. You may post a new question to focus on diagonal $A$. Apr 24, 2021 at 13:09