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We want to take time derivative of

$x^TPx$

and we know $\dot{x}=Ax,~~P^T=P$.

Which one is wrong and why?

1- $d(x^TPx)/dt=\nabla (x^TPx).\dot{x}=2x^TP\dot{x}=2x^TPAx.$

2- $d(x^TPx)/dt=\dot{x}^TPx+x^TP\dot{x}=x^TA^TPx+x^TPAx$

ehsank
  • 415

2 Answers2

1

The second expression is correct, and can be recast into a form that looks more like the first, i.e. $\,\,2\,x^TA^TPx $

You just got your matrix term transposed somehow.

To prevent myself from making transposition mistakes, I usually employ a derivation in terms of the Frobenius product, like so
$$\eqalign{ f &= x^TPx \cr &= P:xx^T \cr\cr df &= P:2\,{\rm sym}(dx\,x^T) \cr &= 2\,{\rm sym}(P):dx\,x^T \cr &= 2\,Px:dx \cr &= 2\,Px:Ax\,dt \cr\cr \frac{df}{dt} &= 2\,Px:Ax \cr &= 2\,A^TP:xx^T \cr &= 2\,x^TA^TPx \cr }$$

where $\,{\rm sym}(M) \equiv \frac{1}{2}(M+M^T)\,$ and $\,A\!:\!B={\rm tr}(A^TB)$ is the Frobenius product.

greg
  • 688
  • you end up with something which is neither 1- nor 2-. For sure $A^TP\neq PA$. – ehsank May 01 '15 at 09:39
  • But notice that ${\rm sym}(A^TP)={\rm sym}(P^TA)={\rm sym}(PA)$, and since the product $x^TPAx$ only uses the symmetric part of $PA$ the products are identical. In other words, $x^TPAx=x^TA^TPx$ – greg May 01 '15 at 12:32
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The second is correct - keyword "chain rule". The first is wrong because it differs.

Hagen von Eitzen
  • 374,180
  • I dont understand why it is wrong, it is simply a chain rule as well. taking the derivative with respect to x, then multipied by the derivative w.r.t time. – ehsank Apr 30 '15 at 11:56