# Find $\operatorname{trace}(BY^{-1})$, given $\mathrm{vec}(Y)=(A\otimes A-I_n)^{-1}\mathrm{vec}(B)$

Given diagonal $$A\in\mathbb{R}^{n\times n}$$, symmetric $$B\in\mathbb{R}^{n\times n}$$, and $$\mathrm{vec}(Y)=(A\otimes A-I_n)^{-1}\mathrm{vec}(B)$$, find the following:

\begin{align} \operatorname{trace}(BY^{-1}). \end{align}

$$A-I>0,$$ $$B$$ is nonsingular.

My attempt:

For special case $$A=aI_n$$:

\begin{align} \mathrm{vec}(Y)=(A\otimes A-I_n)^{-1}\mathrm{vec}(B)=(aI_n\otimes aI_n-I_n)^{-1}\mathrm{vec}(B)=(a^2-1)^{-1}\mathrm{vec}(B). \end{align}

Thus

\begin{align} Y^{-1}=(a^2-1)B^{-1}, \end{align}

and \begin{align} \operatorname{trace}(BY^{-1})=\operatorname{trace}((a^2-1)I_n)=n(a^2-1). \end{align}

For special case $$A=\begin{bmatrix} a_1 & & \\ & a_2 & \\ &&a_2 \end{bmatrix}:$$

\begin{align} \operatorname{trace}(BY^{-1})=a_1^2a_2^2 + a_2^2 - 2. \end{align}

A=[3 0 0; 0 2 0; 0 0 2];
a=[3 2 2]';
T=zeros(1000,1);
for i=1:1000
B=rand(3,2);
[X,K,L] = idare(A,B,zeros(3,3));
T(i)=a'*X.*inv(X)*a;
end
plot(T)


Define the vectors \eqalign{ a = {\rm diag}(A),\quad b = {\rm vec}(B),\quad y = {\rm vec}(Y) \\ } and the matrices \eqalign{ A &= {\rm Diag}(a) \\ M &= A\otimes A - I_n\otimes I_n } Use these to rearrange the given relationship into a matrix equation.
\eqalign{ y &= M^{-1}b \\ b &= My = \left(A\otimes A - I_n\otimes I_n\right)y &= {\rm vec}(AYA - Y) \\ B &= AYA - Y \\ BY^{-1} &= AYAY^{-1} - I_n \\ } The function in question is the trace of that last expression. \eqalign{ \phi &= {\rm Tr}(BY^{-1}) \\&= {\rm Tr}(AYAY^{-1}) - {\rm Tr}(I_n) \\ &= YAY^{-1}:A - n \\ &= {\rm diag}(YAY^{-1}):{\rm diag}(A) - n \\ &= \left(Y^{-T}\odot Y\right)a:a - n \\ &= R_Ya:a - n \\ &= a^TR_Ya - n \\ } The matrix $$R_Y$$ is known as the Relative Gain Array of $$Y$$ and has some interesting properties.
For example, denoting the all-ones vector by $${\tt1},\,$$ it demonstrates Markov-like behavior
$$R_Y{\tt1}={\tt1},\qquad {\tt1}^TR_Y{\tt1}={\tt1}^T{\tt1}=n$$ Re-examining the special case of $$\;\Big(A=\alpha I\iff a=\alpha{\tt1}\Big)\;$$ yields \eqalign{ \phi &= (\alpha{\tt1})^TR_Y(\alpha{\tt1}) - n \\ &= \alpha^2n - n \\ \\ }

In the above, the symbol $$(\odot)$$ denotes the elementwise/Hadamard product
while a colon $$(:)$$ represents the trace/Frobenius product, i.e. \eqalign{ A:B = {\rm Tr}(A^TB) } Finally, the diag() function creates a column vector from the main diagonal of the input matrix, while Diag() creates a diagonal matrix from a vector argument.

### Update

I don't think you're interpreting your simulation correctly.

Here is a short snippet Julia/Matlab code which demonstrates that changing $$B$$ (while holding $$A$$ constant) definitely affects the value of $$\phi$$ and $$R_Y$$
a  = 100*rand(3,1);  A = Diag_(a)
77.2603   0.0      0.0
0.0     83.9703   0.0
0.0      0.0     58.4881

B  = rand(3,3);  B += B';
y  = inv(kron(A,A)-I)*vec(B);  Y = reshape(y,size(B));
Ry = inv(Y').*Y
0.564956   2.99807   -2.56303
2.99807   -0.343684  -1.65439
-2.56303   -1.65439    5.21742

a'*Ry*a
18283.48744321049

B  = rand(3,3);  B += B';
y  = inv(kron(A,A)-I)*vec(B);  Y = reshape(y,size(B));
Ry = inv(Y').*Y
3×3 Array{Float64,2}:
21.8927   17.0631   -37.9558
17.0631    1.88734  -17.9505
-37.9558  -17.9505    56.9063

a'*Ry*a
40704.04750730478

• thanks for detailed answer
– Lee
Jun 15, 2020 at 16:38
• Hi greg, in simulation $R_Y$ doesn't depend on $B$ when $A$ is diagonal. Is it possible to express $R_Y$ in terms of diagonal elements of $A$? For my second special case I found that $R_Y(1,1)=\frac{(a_1a_2-1)^2}{(a_1-a_2)^2}$, but for general case I couldn't figure out
– Lee
Jun 21, 2020 at 10:10
• What does your simulation look like? I tried some code of my own (above) and don't see the effect that you claim.
– greg
Jun 21, 2020 at 15:03
• I fixed diagonal $A$ and for any random $B$ we will get same $a^TR_Ya$, and elements of $A$ are larger than 1. And also $Y$ is solution to DARE, i use the functuon idare(A,B,Q=0). I will include it in my question soon
– Lee
Jun 21, 2020 at 15:32
• I wrote the code from my memory, because my laptop is not with me, but I think should be no bugs.
– Lee
Jun 21, 2020 at 16:00