2

Suppose $P = x^TAx$

How to find $\frac{dP}{dt}$?

if $x' = Bx$ , where $B$ has the same dimension as $A$.

How to find the final answer?

my answer is:

$$\frac{dP}{dt} = 2[(A+A^T)x]x' = 2[(A+A^T)x]Bx$$

However, it seems that $Bx$ is a $(n\times 1)$ vector and $x$ is also an vector so we cannot find the final answer. Is it true?

sleeve chen
  • 8,281

1 Answers1

3

So I suppose $x = x(t)$ and $A$ is a constant vector. Then by product rule

$$P' = (x^TAx)' = (x')^T Ax + x^T Ax' = (Bx)^T Ax + x^TA Bx \ .$$

Not sure how you get $A^T$ and that $2$.

Edit: As suggested by sleeve chan in the comment, the transpose comes from the fact that $(Bx)^T Ax$ is really a function, so

$$(Bx)^T Ax = \big((Bx)^T Ax\big)^T = x^TA^TBx $$

put this back into the original equation you have

$$P' = x^T(A+A^T) Bx\ .$$

You have to be more careful about interchanging the order of your multplication.You can't do that arbitrarily when dealing with matrix.

Still don't know how you get a 2.

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    http://math.stackexchange.com/questions/482742/how-to-calculate-gradient-of-xtax – sleeve chen Apr 23 '14 at 03:11
  • I just find a similar question on the above link; however, it is w.r.t x instead of t. So I think that for example: d(x^2)dt = 2x*(dx/dt), this is why 2 comes from – sleeve chen Apr 23 '14 at 03:14