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?

  • 1
    $\begingroup$ Use \operatorname{trace} and \det. $\endgroup$
    – Invisible
    Jun 11, 2020 at 7:00
  • $\begingroup$ Any information about $A$? $\endgroup$
    – River Li
    Jun 14, 2020 at 4:49
  • $\begingroup$ @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. $\endgroup$
    – Lee
    Jun 14, 2020 at 7:55
  • $\begingroup$ @RiverLi also eigenvalues of $A$ are larger than $1$ $\endgroup$
    – Lee
    Jun 14, 2020 at 7:56
  • $\begingroup$ @Lee I wrote something. $\endgroup$
    – River Li
    Jun 14, 2020 at 9:59

1 Answer 1


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 $\mathrm{Tr}(B^\mathsf{T}XB) = \mathrm{Tr}(B^\mathsf{T}Y^{-1}B)$.

  • $\begingroup$ 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? $\endgroup$
    – Lee
    Apr 24, 2021 at 9:53
  • $\begingroup$ @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 $A$. $\endgroup$
    – River Li
    Apr 24, 2021 at 13:09
  • $\begingroup$ ok, here is it math.stackexchange.com/q/4115464/287178 $\endgroup$
    – Lee
    Apr 25, 2021 at 4:43

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