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For a given $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{n \times m}$, where $(A,B)$ is controllable

$$\begin{array}{ll} \underset{X \in \mathbb{R}^{n\times n}}{\text{minimize}} & \mathrm{tr} \left( B' X B \right)\\ \text{subject to} & X = A'XA - A'XB(B'XB + I)^{-1}B'XA\end{array}$$

For a specific $A$ and $B$, I can use Matlab to solve the DARE (discrete-time algebraic Riccati equation) and insert the resulting $X$ to find $\mathrm{tr}(B'XB)$. Is it possible to get the general answer in terms of $A$ and $B$?

EDIT:

Lagrangian:

$$L(X, \Lambda) := {\rm Tr}\left(B^T XB \right) - {\rm Tr}\left(\Lambda^T \left[ X - A^TXA + A^TXB \left( B^TXB + I \right)^{-1} B^T X A \right] \right)$$

Taking the gradient:

$$\frac{\partial{L(X, \Lambda)}}{\partial{X}}=BB'-\Lambda+A\Lambda A'-\frac{\partial{\mathrm{tr}\Lambda A'XB(B'XB+I)^{-1}B'XA}}{\partial{X}}$$

FURTHER EDIT: I think $X$ is symmetric, then

$$\begin{align} \frac{\partial{L(X, \Lambda)}}{\partial{X}}&=BB'-\Lambda+A\Lambda A'- A \Lambda \left( B M^{-1} B^T X^T A \right)^T \\&+ \left( A^T X B M^{-1} B^T \right)^T \Lambda \left( B M^{-1} B^T X^T A \right)^T - A \Lambda^T \left( B M^{-1} B^T X^T A \right)^T\\&=BB'-\Lambda+(I-B(B'XB+I)^{-1}B'X)A\Lambda A'(I-XB(B'XB+I)^{-1}B') \end{align}$$

Equating the last expression to zero, we get Lyapunov equation:

\begin{equation} \Bigg[(I-B(B'XB+I)^{-1}B'X)A\Bigg]\Lambda \Bigg[A'(I-XB(B'XB+I)^{-1}B')\Bigg]-\Lambda+\Bigg[BB'\Bigg]=0. \end{equation}

Solving for $\Lambda$, we get $\Lambda=\sum_{k=0}^\infty\Big[(I-B(B'XB+I)^{-1}B'X)A\Big]^kBB'\Big[(I-B(B'XB+I)^{-1}B'X)A\Big]^{k^*}$.

Any tips how I can continue?

Lee
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  • What is "DARE"? – user550103 Jun 01 '20 at 11:51
  • @user550103 discrete time algebraic Riccati equation – Lee Jun 01 '20 at 11:52
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    Lagrangian can be constructed as: $L(X, \Lambda) := {\rm Tr}\left(B^T XB \right) - {\rm Tr}\left(\Lambda^T \left[ X - A^TXA + A^TXB \left( B^TXB + I \right)^{-1} B^T X^T A \right] \right) $ – user550103 Jun 01 '20 at 11:57
  • @RodrigodeAzevedo do you know how to find $\frac{\partial{\mathrm{tr}\Lambda A'XB(B'XB+I)^{-1}B'XA}}{\partial{X}}$? – Lee Jun 01 '20 at 13:36
  • @thanks, I made an edit, do you know how to find the last derivative? – Lee Jun 01 '20 at 13:37
  • Shouldn't that DARE (in most if not all cases) only have a finite number of solutions and thus your cost function could just be checked for each case in order to find the minimum? – Kwin van der Veen Jun 02 '20 at 01:57

1 Answers1

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Since you asked for the gradient of one of the components of Lagrangian, here we go.

Let us define the Frobenius product by a colon, for brevity, i.e., \begin{align} {\rm Tr}\left( A^T B C \right) := A: BC \end{align}

We will use the cyclic property of trace, e.g., \begin{align} A: BCD = B^T A: CD = B^TAD^T: C \end{align}

Let us define for simplicity, \begin{align} M^{-1} := \left( B^T X B + I \right)^{-1} \end{align} And the corresponding differential which will be handy later. \begin{align} dM^{-1} = -M^{-1} dM M^{-1} = -M^{-1} B^T dX B M^{-1} \end{align}

Now the part that you are interested to find the gradient can be rewritten as \begin{align} \phi &:= {\rm Tr}\left( \Lambda^T \left[ A^T X B \left( B^T X B + I \right)^{-1} B^T X^T A \right] \right) \\ &= \Lambda : A^T X B M^{-1} B^T X^T A \end{align}

Compute the differential and then gradient. \begin{align} d\phi &= \left\{ \Lambda : A^T dX B M^{-1} B^T X^T A + A^T X B dM^{-1} B^T X^T A + A^T X B M^{-1} B^T dX^T A \right\} \\ &= \left\{ \Lambda : A^T dX B M^{-1} B^T X^T A - A^T X B M^{-1} B^T dX B M^{-1} B^T X^T A + A^T X B M^{-1} B^T dX^T A \right\} \\ &= \left\{ \left[ A \Lambda \left( B M^{-1} B^T X^T A \right)^T: dX \right] - \left[ \left( A^T X B M^{-1} B^T \right)^T \Lambda \left( B M^{-1} B^T X^T A \right)^T : dX \right] + \left[ A \Lambda^T \left( B M^{-1} B^T X^T A \right)^T : dX \right] \right\} \end{align}

The gradient is \begin{align} \frac{\partial \phi}{\partial X} &= A \Lambda \left( B M^{-1} B^T X^T A \right)^T - \left( A^T X B M^{-1} B^T \right)^T \Lambda \left( B M^{-1} B^T X^T A \right)^T + A \Lambda^T \left( B M^{-1} B^T X^T A \right)^T \end{align}

You can simplify further. I hope this helps

Lee
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user550103
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