# Prove that $\mathrm{Tr}(B^\mathsf{T}Y^{-1}B)$ is independent of $B$

Given diagonal $$A\in\mathbb{R}^{n\times n}$$ with all eigenvalues larger than $$1$$, and minimal polynomial $$\alpha(\lambda)$$.

Matrix is called cyclic if its minimal polynomial is equal to characteristic polynomial.

Here $$A=\begin{bmatrix} A_1 & \\ & A_2 \end{bmatrix}$$, where $$A_i$$, for $$i=1,2,$$ are cyclic with minimal polynomials $$\alpha_i(\lambda)$$, such that $$\alpha_1(\lambda)=\alpha(\lambda)$$ and $$\alpha_2(\lambda)$$ divides $$\alpha_1(\lambda)$$. For example: $$A=\mathrm{diag}(5,4,3,2,3,2)$$, here $$A_1=\begin{bmatrix} 5 & &&\\ & 4&&\\ & &3&\\ & &&2 \end{bmatrix}$$ and $$A_2=\begin{bmatrix} 3 & \\ & 2 \end{bmatrix}$$.

Basically, $$A_2$$ collects all repeating diagonal elements, but $$A$$ cannot have more that $$2$$ same diagonal elements. $$A=\mathrm{diag}(2,2,2)$$ is not possible. And we assume $$\mathrm{rank}[\lambda I-A \quad B]=n$$ for all $$\lambda\in\mathbb{R^+}$$, i.e. $$(A,B)$$ is stabilizable.

If $$BB^\mathsf{T}=AYA-Y$$ where $$B\in\mathbb{R}^{n\times2}$$, prove that $$\mathrm{Tr}(B^\mathsf{T}Y^{-1}B)$$ is independent of $$B$$.

My attempt:

For special case when all eigenvalues of $$A$$ are equal to $$a$$, we have $$\mathrm{vec}(BB^\mathsf{T}) = (A \otimes A - I)\mathrm{vec}(Y)$$, where $$\otimes$$ denote the Kronecker product. And since $$(A \otimes A - I)$$ is nonsingular, we have

\begin{align} \mathrm{vec}(Y) &= (A \otimes A - I)^{-1}\mathrm{vec}(BB^\mathsf{T})\\ &=\frac{1}{a^2-1}\mathrm{vec}(BB^\mathsf{T})\\ \end{align}

Which means $$Y=\frac{1}{a^2-1}BB^\mathsf{T}$$ and $$\mathrm{Tr}(B^\mathsf{T}Y^{-1}B)=\mathrm{Tr}(BB^\mathsf{T}Y^{-1})=\mathrm{Tr}(BB^\mathsf{T}(a^2-1)(BB^\mathsf{T})^{-1})=2(a^2-1).$$

I think $$BB^\mathsf{T}$$ is singular, but maybe we can take pseudo-inverse.

Conjecture: For general case

\begin{align} \mathrm{Tr}(B^\mathsf{T}Y^{-1}B)=\mathrm{det}(A_1)+\mathrm{det}(A_2)-2, \end{align}

Matlab simulation agrees with the conjecture.

I have edited the question. Since this math problem arises from engineering problem, I think my initial question was not clear for mathematical audience. I hope now it is self-containing.

• Comments are not for extended discussion; this conversation has been moved to chat. May 1, 2021 at 18:47