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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    May 1, 2021 at 18:47


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