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Given $A=\begin{bmatrix} a_1 & & \\ & a_2 & \\ &&a_2 \end{bmatrix}$ and $B\in\mathbb{R}^{3\times2}$, where $a_1,a_2>1$, and $X=AX(I+BB'X)^{-1}A$.

1) Find $\mathrm{tr}(AX^{-1}AX)$ in terms of $a_i$.

2) Show that $\frac{x_1\cdot\begin{vmatrix} x_2 & x_{23}\\ x_{23}&x_3 \end{vmatrix}}{|X|}$ is independent of $B$, where $|X|=\mathrm{det}X$.


My attempt: $X$ is the solution to discrete-time algebraic Riccati equation.

Let $X=\begin{bmatrix} x_1 & x_{12} & x_{13}\\ x_{12} & x_2 & x_{23}\\ x_{13}&x_{23}&x_3 \end{bmatrix}$ and $z=x_1\cdot\begin{vmatrix} x_2 & x_{23}\\ x_{23}&x_3 \end{vmatrix}$, then I found that

\begin{align} \mathrm{tr}(AX^{-1}AX) =(a_1-a_2)^2\frac{z}{|X|}+(2a_1a_2+a_2^2). \end{align}

I did simulation by fixing $A$ and found out that $\frac{z}{|X|}$ doesn't depend on $B$, thus $\mathrm{tr}(AX^{-1}AX)$ also doesn't depend on $B$ (as long as $B$ is full rank).


Observation 1: $A=\begin{bmatrix} 7 & & \\ & 2 & \\ &&2 \end{bmatrix}$ and $A=\begin{bmatrix} 2 & & \\ & 7 & \\ &&7 \end{bmatrix}$ has equal $\frac{z}{|X|}$.

Observation 2: The larger the gap between $a_1$ and $a_2$, the smaller $\frac{z}{|X|}$ and vice-versa.

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  • $\begingroup$ what are $B'$ and $A'$? $\endgroup$
    – Quillo
    Jun 4, 2020 at 17:35
  • 1
    $\begingroup$ @mrcntn Matrix transpose $\endgroup$
    – Lee
    Jun 4, 2020 at 17:56
  • $\begingroup$ Forgive me if I am mistaken, but isn't $\text{tr}(AX^{-1}A'X) = \text{tr}(A'A)$ since the matrices inside the trace can commute? $\endgroup$
    – Theorem
    Jun 4, 2020 at 18:10
  • $\begingroup$ @Theorem They cannot commute like that, they should preserve the order, $\mathrm{tr}(AX^{-1}A'X)=\mathrm{tr}(X^{-1}A'XA)=\mathrm{tr}(A'XAX^{-1})=\mathrm{tr}(XAX^{-1}A')$ $\endgroup$
    – Lee
    Jun 4, 2020 at 18:19

1 Answer 1

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Assume that $\det X \ne 0$.

From $X = AX(I + BB^\mathsf{T}X)^{-1}A$, we have $I + BB^\mathsf{T}X = AX^{-1}AX$ and $BB^\mathsf{T} = AX^{-1}A - X^{-1}$. Let $Y = X^{-1}$. We have $BB^\mathsf{T} = AYA - Y$. Let $$Y=\begin{bmatrix} y_1 & y_{12} & y_{13}\\ y_{12} & y_2 & y_{23}\\ y_{13}&y_{23}&y_3 \end{bmatrix}, \quad B = \begin{bmatrix} b_1 & b_2 \\ b_3 & b_4 \\ b_5 & b_6 \end{bmatrix}. $$ From $BB^\mathsf{T} = AYA - Y$, we solve $Y$ uniquely \begin{align} y_1 &= \frac{b_1^2 + b_2^2}{a_1^2 - 1}, \quad y_{12} = \frac{b_1b_3 + b_2b_4}{a_1a_2 - 1}, \quad y_{13} = \frac{b_1b_5 + b_2 b_6}{a_1a_2 - 1}, \\ y_2 &= \frac{b_3^2 + b_4^2}{a_2^2 - 1}, \quad y_{23} = \frac{b_3b_5 + b_4b_6}{a_2^2-1}, \quad y_3 = \frac{b_5^2+b_6^2}{a_2^2-1}. \end{align}

By using the well-known relation $M^{-1} = \frac{1}{\det M} \mathrm{adj}(M)$, from $Y = X^{-1}$ and $X = Y^{-1}$, we have \begin{align} y_1 &= \frac{\begin{vmatrix} x_2 & x_{23}\\ x_{23}&x_3 \end{vmatrix}}{\det X}, \\ x_1 &= \frac{\begin{vmatrix} y_2 & y_{23}\\ y_{23}&y_3 \end{vmatrix}}{\det Y}. \end{align} Thus, we have $$\frac{z}{\det X} = y_1x_1 = y_1 \frac{\begin{vmatrix} y_2 & y_{23}\\ y_{23}&y_3 \end{vmatrix}}{\det Y} = \frac{(a_1a_2-1)^2}{(a_1-a_2)^2}.$$ Then, we have $$\mathrm{Tr}(AX^{-1}AX) = (a_1-a_2)^2\frac{z}{\det X}+(2a_1a_2+a_2^2) = a_1^2a_2^2 + a_2^2 + 1.$$

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