Is, by any chance, these two formulas equivalent? x is a complex vector, and H means the complex transposition operation. I should get from the first to the second. But it seems to me that this is not equivalent.
$ \begin{align} \dot x & = \left( \mathbb{I} - A^{-T}\frac{x x^H}{x^H A^{-T} x} \right) A \, x, \end{align}$
$\dot x = (\mathbb{I} - \frac{x x^H} {x^H x} )Ax \nonumber $
The second expression is obtained from the difference between the complex nx1 vector $Ax$ and the projection of $Ax$ onto x.
$\dot x = A x - proj_x Ax$
Thanks
Edit: The context is the following, I have a dynamical system $\dot x = Ax $ that needs constraint solutions to the unit circle. Therefore, $x^ H x = 1$. The first equation was obtained by constraining the system using a lagrangian multiplier. The second is obtained by simply projecting $\dot x = Ax$ into the tangent direction of x (therefore, the motion is trapped into a circular motion of radius 1.). These expressions should be equivalent.
Define the matrices $$\eqalign{ \def\c#1{\color{red}{#1}} \def\A{A^{-T}} \def\x{x^H} P &= \frac{\A x\x A}{\x\A x},\qquad Q = \frac{x\x A}{\x x} \\ }$$ Note that $P\ne Q$ (unless $A$ equals the identity matrix), however $$\eqalign{ \x P &= \x A,\qquad \x Q = \x A \\ }$$ The proposed equality reduces to $$\eqalign{ \dot x \;=\; (A-P)x &\;\overset{?}{=}\; (A-Q)x \\ Px &\;\overset{?}{=}\; Qx \\ P &\;\overset{?}{=}\; Q \\ }$$ which is false. However, the Differential Equation (DE) can be multiplied to obtain $$\eqalign{ \x\dot x &= \x Ax - \c{\x P}x \\ &= \x Ax - \c{\x Q}x \\ &= \x Ax - \c{\x A}x \\ &= 0 \\ }$$ When $x$ is $\sf Real$, this is equivalent to the constraint $x^Tx=\tt1\,$ and adds nothing new.
When $x$ is $\sf Complex$, the constraint only says $(\x\dot x = -\dot{x}^H x)$ and it is the DE which zeroes them.