3
$\begingroup$

The projection onto a parametrised vector $v(\lambda)$ is $P_v = \frac{vv^{T}}{v^{T}v}.$ Its complement is $$P = I-\frac{vv^T}{v^{T}v}.$$ I've got an expression containing this complementary projection and I need its derivative. How do I calculate

$$\frac{\partial P(v(\lambda))}{\partial \lambda} \text{ ?}$$

I started with $$\cfrac{\partial P(v(\lambda))}{\partial \lambda} = \cfrac{\partial P(v(\lambda))}{\partial v} \cfrac{\partial v(\lambda))}{\partial \lambda}$$ where only the expression $\cfrac{\partial vv^{T}}{\partial v}$ I can't handle. How can I find this derivative of a matrix with respect to a vector, or the original derivative with respect to the scalar parameter $\lambda$?

$\endgroup$
6
  • 3
    $\begingroup$ I think it is simpler to compute the derivative in coordinates. Since $vv^T=[v_iv_j]$ one differentiates with respect to $\lambda$ and finds $\partial_\lambda P = v' v^T + v(v')^T$. $\endgroup$ Jan 30, 2016 at 19:43
  • $\begingroup$ (It may also help to remember that transposition commutes with differentiation.) $\endgroup$
    – Neal
    Jan 30, 2016 at 20:26
  • $\begingroup$ Which makes it: $$\frac{\partial P(v(\lambda))}{\partial \lambda} = - \frac{(v^{T}v)(v'v^{T}+v(v')^{T}) -vv^{T} ((v')^{T}v+v^{T}v') }{||v||^{4}} $$, correct? $\endgroup$
    – mike
    Jan 30, 2016 at 20:38
  • $\begingroup$ @mike Giuseppe's answer is correct. I'm not sure what your comment is all about $\endgroup$ Jan 31, 2016 at 2:15
  • $\begingroup$ @mike oh! You're applying the quotient rule to your normalized version. Yes, you did so correctly. Note that what you've written only makes sense if you treat $v^Tv$ as a scalar (in other words, the matrices are not conformable if you regroup that product). $\endgroup$ Jan 31, 2016 at 2:19

2 Answers 2

1
$\begingroup$

Thanks to Giuseppe Negro I found the answer to be:

$$\frac{\partial P(v(\lambda))}{\partial \lambda} = - \frac{(v^{T}v)(v'v^{T}+v(v')^{T}) -vv^{T} ((v')^{T}v+v^{T}v') }{||v||^{4}}$$

after using the quotient rule. Remark: I used the general case here, where $v$ is not necessarily of unit length, so it needs to be normalised by its norm.

$\endgroup$
1
$\begingroup$

Multiply to clear the fraction $$\eqalign{ (v^Tv)\,P &= (v^Tv)\,I - (vv^T) \cr }$$ Differentiate (using $d=\frac{d}{d\lambda}$ for ease of typing) $$\eqalign{ d(v^Tv)\,P + (v^Tv)\,dP &= d(v^Tv)\,I - d(vv^T) \cr (dv^T\,v+v^Tdv)\,P + (v^Tv)\,dP &= (dv^T\,v+v^Tdv)\,I - (dv\,v^T+v\,dv^T) \cr (v^Tv)\,dP &= (dv^T\,v+v^Tdv)\,(I-P) - (dv\,v^T+v\,dv^T) \cr }$$ Solve for $dP$ $$\eqalign{ dP &= \frac{(dv^T\,v+v^Tdv)\,(I-P) - (dv\,v^T+v\,dv^T)}{v^Tv} \cr &= \frac{(dv^T\,v+v^Tdv)\,(vv^T) - (v^Tv)(dv\,v^T+v\,dv^T)}{(v^Tv)^2} \cr }$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.