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We usually consider $\dot{x} = f(x)$, where $x$ is a vector.

Now, I want to consider $$\dot{X}=f(X,U),$$ where $X$ is a square matrix $\mathbb{R}^{n\times n}$ state, $U$ is a square matrix variable $\mathbb{R}^{n\times n}$, like a control input. If I want to consider its stability, I have to use Lyapunov stability theorem, $$V(X)>0,$$ for $X\neq 0$, And $$\dot{V}<0$$along $\dot{X}=f(X)$. One way to do this i let $\dot{X}=-\nabla V$, so $$\dot{V} = \langle\nabla V, \dot{X}\rangle=-\langle\nabla V, \nabla V\rangle = -\|\nabla V\|_F^2.$$ So I have to let $$f(X,U) = -\nabla V$$ to solve for $U$. However, it is not the case of vector, we cannot easily "make them equal", for example, they may not have the same rank.

------------------------------new idea-----------------------------

To relax such condition, I would consider the set $$\mathcal{C} = \{W : \langle W,-\nabla V \rangle \geq 0\},$$ which implies $W\in \mathcal{C}$ points to the descent direction if $V$ is a convex function in $Q$. So instead of letting $f(X,U) = -\nabla V$, let $f(X,U) = W$, $W\in \mathcal{C}$. So now I probably have more flexible choice.

Any reference or paper about the matrix version of dynamic system / stability analysis / Lyapunov theory?

Sincerely appreciate this help.

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    $\begingroup$ Is there some additional structure or restrictions to this problem? For example, if $X$ is a real $n \times n$ matrix, is there anything wrong with treating it as a vector in $ \mathbb{R}^{n\times n}$? Usually with Lyapunov functions, one starts with the vector field $f$ and then one looks for an appropriate Lyapunov function. However here it looks like you might have the freedom to adjust your function $f$? $\endgroup$
    – Jonathan J.
    Commented Jan 19, 2021 at 20:22
  • $\begingroup$ @JonathanJ. Thanks for good catch. I modify my question a little bit. $\endgroup$
    – Denny
    Commented Jan 20, 2021 at 9:49
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    $\begingroup$ So far nothing in the problem makes this a particular matrix problem. If I understand correctly, we have a system $\dot x = f(x,u)$ on a vector space (the vectors happen to be matrices). Now find a smooth Lyapunov function $V$ such that you can choose $u$ in dependence of $x$ such that $\dot V$ becomes negative definite. First of all, already in the vector case this is not always easy (as your wording seems to suggest). What you are looking for is called a control Lyapunov function (clf) and maybe you first read up on those. BTW: the condition $f(x,u)= -\nabla V(x)$ is quite restrictive. $\endgroup$
    – Fabian Wirth
    Commented Jan 20, 2021 at 16:22
  • $\begingroup$ @FabianWirth I will consider the set $\mathcal{C} = \{v: \langle v, -\nabla V(x) \rangle \geq 0\}$. If $V$ is convex, then $\mathcal{C}$ is the relaxation of such constraint $f = -\nabla V(x)$. So I can choose anything in $\mathcal{C}$. $\endgroup$
    – Denny
    Commented Jan 25, 2021 at 17:38

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