Finding deriviative of lyapunov's equation (chain rule/linear algebra)

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SUMMARY

This discussion focuses on deriving the derivative of Lyapunov's equation using linear algebra and the chain rule. The Lyapunov function is defined as V(x) = xTP x, where P is a symmetric positive-definite matrix. The derivative is expressed as d/dt V(x) = xTFTP x + xTP F x, with F representing the state matrix. The discussion highlights the importance of understanding matrix transposition properties in this derivation.

PREREQUISITES
  • Understanding of linear algebra concepts, specifically matrix operations.
  • Familiarity with Lyapunov functions and their applications in control theory.
  • Knowledge of the chain rule in differentiation.
  • Basic understanding of state-space representation in dynamic systems.
NEXT STEPS
  • Study the properties of symmetric positive-definite matrices in linear algebra.
  • Learn about the application of Lyapunov's stability theorem in control systems.
  • Explore matrix calculus, particularly the differentiation of matrix functions.
  • Investigate the state-space representation and its implications in dynamic system analysis.
USEFUL FOR

This discussion is beneficial for students and professionals in control systems engineering, particularly those working with Lyapunov stability analysis and linear algebra applications in dynamic systems.

james1234
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Hi there!

I am just trying to work though a simple deriviation presented in a txt book (feedback control of dynamic systems p 616). I am REALLY new to linear algebra and they have skipped too many steps!

they have used the chain rule for differentiation of V(x) (a lyapunov function) where V = xTPx ; where P is a symetric positive matrix (i have not been able to find reference to a 'positive matrix'.. is a positive-definite matrix the same thing?)

Thus we have;

d/dtxTPx = x\dot{}TPx + xTPx\dot{}

from there we somehow get to

xT(FTP + PF)x (where i believe F is the 'state' matrix usually represened as A ie X dot = Fx + bu )

if someone could point me in the right direction it would really be appreciated!
 
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It doesn't have much to do with calculus, it's just plugging in the definition.
The only step that you might be missing is that (AB...XYZ)T = ZT YT XT ... BT AT for some matrices / vectors A, ..., Z.
 
hi CompuChip

thankyou so much for for taking the time to reply!
despite all the time I have spent pooring over it I just couldn't see it!
(and yes I was missing that step!)

thanks again! :)
 

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